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Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but second is not an extreme value. Comment on particular cases when (i) ${a+b+c=0}$, (ii) ${a=b=c}$.
My Attempt:
$${F=f+\lambda \phi =x^3 + y^3 + z^3 - 3xyz + 3\lambda(ax + by + cz-1)}$$
$${\frac13F_x = x^2-yz+\lambda a = 0}{\text{ ...(1)}}$$
$${\frac13F_y = y^2-xz+\lambda b = 0}{\text{ ...(2)}}$$
$${\frac13F_z = z^2-xy+\lambda c = 0{\text{ ...(3)}}}$$
$${(1)x+(2)y+(3)z \implies f+\lambda (1) = 0 \implies \lambda = -f}$$
$${(1)+(2)+(3) \implies}{x^2+y^2+z^2-xy-yz-zx=(a+b+c)f}$$
$${\implies f/(x+y+z)=(a+b+c)f}\,,$$
then ${f=0}$ or ${x+y+z=1/(a+b+c)}$.
Also, ${(1)-(2) \implies x^2-y^2-z(x-y)=f(a-b) \implies \frac{x-y}{a-b}=f\frac{a+b+c}{x+y+z}}$
Similarly ${(2)-(3)}$ and ${(3)-(1)}$, then we get
${\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}=f\frac{a+b+c}{x+y+z}}$
I don't know how to proceed further. I couldn't get the other stationary value of ${f}$. I need help in proceeding further to calculate stationary points and/or stationary values, if at all, my method is correct.
| Here I shall find all critical points without verifying what kind of critical points they are (which seems to be your question). My notations are similar to yours, but a bit different, so I shall work from scratch but my answer borrows a lot of your ideas (great attempt, by the way). However, I do not expect that any optimizing point will be a global one. Therefore, be careful and do not assume that any optimizing point will yield a global optimum.
For $X,Y,Z\in\mathbb{R}$, let $$\begin{align}F(X,Y,Z)&:=X^3+Y^3+Z^3-3XYZ\\&=(X+Y+Z)(X^2+Y^2+Z^2-YZ-ZX-XY)\end{align}$$
and $$G(X,Y,Z):=aX+bY+cZ-1\,,$$
where $a$, $b$, and $c$ are fixed real numbers. The task is as follows:
$$\begin{array}{ll}\text{optimize}&F(X,Y,Z) \\\text{subject to} & X,Y,Z\in\mathbb{R}\\&G(X,Y,Z)=0\,.\end{array}$$
We set up the Lagrangian $\mathcal{L}(X,Y,Z,\Lambda)$ for $X,Y,Z,\Lambda\in\mathbb{R}$ by
$$\mathcal{L}(X,Y,Z,\Lambda):=F(X,Y,Z)-3\,\Lambda\,G(X,Y,Z)\,.$$
If $(x,y,z)\in\mathbb{R}^3$ is such that $(X,Y,Z):=(x,y,z)$ is a solution to this optimization problem, then there exists $\lambda\in\mathbb{R}$ for which
$$\frac{\partial \mathcal{L}}{\partial V}(x,y,z,\lambda)=0$$
for all variables $V\in\{X,Y,Z,\Lambda\}$. That is, we have the following equations:
$$x^2-yz=\lambda\,a\,,\tag{1}$$
$$y^2-zx=\lambda\,b\,,\tag{2}$$
$$z^2-xy=\lambda\,c\,,\tag{3}$$
and
$$ax+by+cz=1\,.\tag{4}$$
We let $f:=F(x,y,z)$. Adding $x$ times (1), $y$ times (2), and $z$ times (3) yields
$$\begin{align}f&=x(x^2-yz)+y(y^2-zx)+z(z^2-xy)=x(\lambda\,a)+y(\lambda\,b)+z(\lambda\,c)\\&=\lambda\,(ax+by+cz)=\lambda\cdot 1=\lambda\,,\end{align}$$ due to (4). By adding (1), (2), and (3), we obtain
$$\begin{align}
f&=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)
\\&=(x+y+z)\big((x^2-yz)+(y^2-zx)+(z^2-xy)\big)
\\&=(x+y+z)\big(\lambda\,a+\lambda\,b+\lambda\,c)=(x+y+z)(a+b+c)\lambda\\&=(x+y+z)(a+b+c)f\,.
\end{align}$$
This means $(x+y+z)(a+b+c)=1$ or $f=0$.
Case I: $a+b+c=0$. Then, $f=0$ must hold (whence $\lambda=f=0$). By adding (1), (2), and (3) together, we obtain
$$\frac{(y-z)^2+(z-x)^2+(x-y)^2}{2}=(x^2-yz)+(y^2-zx)+(z^2-xy)=0\,.$$
Thus, $x=y=z$ must be the case. Ergo,
$$1=ax+by+cz=ax+bx+cx=(a+b+c)x=0\,,$$
which is a contradiction. Consequently, there does not exist a critical point when $a+b+c=0$.
Case II: $a+b+c\neq 0$ but $f=0$. By adding (1), (2), and (3) together, we conclude, as in Case I, that $x=y=z$. Ergo,
$$1=ax+by+cz=ax+bx+cx=(a+b+c)x\text{ implies }x=y=z=\frac{1}{a+b+c}\,.$$
This yields
$$(f,x,y,z)=\left(0,\frac1{a+b+c},\frac1{a+b+c},\frac1{a+b+c}\right)\,.\tag{5}$$
We shall see later that, when $a=b=c$, then this is the only case that yields a critical point.
Case III: $a+b+c\neq 0$ and $f\neq 0$. Then, we must have $x+y+z=\dfrac1{a+b+c}$. Subtracting (2) from (1) gives us $(x-y)(x+y+z)=\lambda(a-b)=f(a-b)$, so that
$$x-y=(a+b+c)f(a-b)\,.$$
Similarly,
$$y-z=(a+b+c)f(b-c)$$
and
$$z-x=(a+b+c)f(c-a)\,.$$
Set $k:=(a+b+c)fa-x$. Then, we get
$$x=(a+b+c)fa-k\,,\,\,y=(a+b+c)fb-k\,,\text{ and }z=(a+b+c)fc-k\,.$$
As $x+y+z=\dfrac1{a+b+c}$, we obtain
$$(a+b+c)^2f-3k=\frac{1}{a+b+c}\,.$$
Because $ax+by+cz=1$, we must have
$$(a+b+c)(a^2+b^2+c^2)f-(a+b+c)k=1\,.$$
This shows that
$$\begin{align}&(a^3+b^3+c^3-3abc)f\\&\phantom{aaa}=\frac{3\big((a+b+c)(a^2+b^2+c^2)f-(a+b+c)k\big)-(a+b+c)\big((a+b+c)^2f-3k\big)}{2}\\&\phantom{aaa}=\frac{3\cdot 1-(a+b+c)\cdot\left(\frac{1}{a+b+c}\right)}{2}=1\,.\end{align}$$
Consequently, $a^3+b^3+c^3-3abc\neq 0$, implying that $a$, $b$, and $c$ are not all equal, and so
$$f=\frac{1}{a^3+b^3+c^3-3abc}\,,\text{ which leads to }k=\frac{bc+ca+ab}{a^3+b^3+c^3-3abc}\,.$$ Thence,
$$\begin{align}(f,x,y,z)&=\Biggl(\frac{1}{a^3+b^3+c^3-3abc},\frac{a^2-bc}{a^3+b^3+c^3-3abc}\\&\phantom{aaaaa},\frac{b^2-ca}{a^3+b^3+c^3-3abc},\frac{c^2-ab}{a^3+b^3+c^3-3abc}\Biggr)\,.\end{align}\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$ The derivative of the denominator is $\cos(x)$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$
Is that right so far?
Thanks for the help in advance.
Best Regards,
Ahmed Hossam
| As we know that $$\lim_{x\to 0}{x\over \sin x}=1$$therefore $$\lim_{x\to 0}{x^2\cos {1\over x}\over \sin x}=\lim_{x\to 0}x\cos{1\over x}=\lim_{u\to \infty}{\cos u\over u}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x +1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2 - x +2}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x \left( x - 1 +\frac{2}{x}\right)}{x \left( \sqrt{1+\frac{1}{x}}+1-\frac{1}{x} \right)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1}{2} = \infty
\end{equation}
Edit: Added correct steps for completeness, thanks for the quick answers!
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x^2+2x -1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{2x}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\frac{2}{1+1} = 1
\end{equation}
| Set $1/x=h\implies h\to0^+$
and $\sqrt{x^2+1}=\dfrac{\sqrt{1+h^2}}{|h|}=\dfrac{\sqrt{1+h^2}}h$ as $h>0$ as $h\to0^+$
So, we have $$ \lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}$$
$$=1+\lim_{h\to0^+}\dfrac{\sqrt{1+h^2}-1}h$$
$$=1+\lim_{h\to0^+}\dfrac{1+h^2-1}h\cdot\lim_{h\to0^+}\dfrac1{\sqrt{1+h^2}+1}$$
$$=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How many sequences can be made with 5 digits so that the difference between any two consecutive digits is $1$? Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten-digit sequences can be written so that the difference between any two consecutive digits is $1$?
I was wondering if my solution is right.
Let $a(n)$ be the number of $n$ digit sequences that end with $0$ or $4$ so that the difference between any two consecutive digits is $1$.
$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so that the difference between any two consecutive digits is $1$.
$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.
$x(n)$ be the number of n digit sequences so that the difference between any two consecutive digits is $1$.
$x(n) = a(n) + b(n) + c(n)$
$a(n) = b(n-1)$
$b(n) = a(n-1) + 2c(n-1)$
$c(n) = b(n-1)$
By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$
We know that $b(1) = 2, b(2) = 4$.
The characteristic equation for this recursion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.
We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.
| Your approach is correct, with the above 2 relations mentioned by you:
$$a(n) = b(n-1) = c(n)$$
$$b(n) = a(n-1) + 2c(n-1)$$
we can easily get the relation $$x(n)=3x(n-2)$$
Hence $x(10) = 3x(8) = 3^2 x(6) =...= 3^4 x(2)$, where $x(2) = 8$.
So, the answer to your question would be $2^3 3^4 = \boldsymbol{648}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044871",
"timestamp": "2023-03-29T00:00:00",
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Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the book this is the answer only for $z=0$. How should I find the answers for the other two? Thanks.
| Hints :
For the $z=1$ case :
You need to create terms of the form $z-1$. You can manipulate your fraction decomposition that you already carried out, as :
$$f(z) = \frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z} = \frac{1}{1+(z-1)} + \frac{1}{1-z} $$
$$=$$
$$\frac{1}{(z-1)\left(\frac{1}{z-1} + 1\right)} - \frac{1}{z-1} = \frac{1}{z-1}\left(\frac{1}{\frac{1}{z-1} + 1}\right)$$
Now, recall the geometric series $\frac{1}{1+w} = \sum_{n=1}^\infty (-1)^nw^n$. Let $w = \frac{1}{z-1}$. Thus :
$$f(z) = \frac{1}{z-1}\sum_{n=0}^\infty (-1)^n \left(\frac{1}{z-1}\right)^n =\sum_{n=1}^\infty (-1)^{n-1}\left(\frac1{z-1}\right)^{n+1}$$
For the $\infty$ case :
Recall the geometric series $\frac{1}{1-w} = \sum_{n=1}^\infty w^{n}$ when $|w| <1$. Thus, for $|z| > 1$, we can write :
$$f(z) = \frac{1}{z(1-z)}= -\frac{1}{z^2(1-\frac{1}{z})}=-\sum_{n=0}^{\infty}z^{-n-2}$$
Alternative : Let $w = 1/z$ and calculate the Laurent Series for $w =0$ which happens when $z \to \infty$.
| {
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"url": "https://math.stackexchange.com/questions/3045468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the other asymptote of $y=\sqrt{x^2+x}$ The task is to find the asymptotes of $y=\sqrt{x^2+x}$.
I first calculated the limits to infinity and found that $\lim_{x \to \pm}y= \infty$.
Next, to find $m_{1,2}$: $$m_1=\lim_{x \to +\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_1$$
and $$m_2=\lim_{x \to -\infty}\frac{y}{x}=\frac{\sqrt{x^2+x}}{x}=\sqrt{1+ \frac{1}{x}}=1=m_2$$
Now to find $c_{1,2}$: $$c_1=\lim_{x \to +\infty}y-mx={\sqrt{x^2+x} - x}=\frac{(\sqrt{x^2+x} - x)({\sqrt{x^2+x} +x})}{{\sqrt{x^2+x} +x}}=\frac{x}{{\sqrt{x^2+x} +x}}= \frac{x}{x+{|x|\sqrt{1+\frac{1}{x}}}} \to \frac{x}{x+{|x|}}
\approx \frac{1}{1+\frac{|x|}{x}}=\frac{1}{2} $$
$$c_2=\frac{1}{0}?$$
From here, it is clear that as $x$ tends to positive infinity, we have $c_1=1/2$, therefore, $y=\frac{1}{2}(2x+1)$ is an asymptote. However, when I try to calculate the limit to negative infinity, I have an indeterminate $\frac{1}{0}$. However, according to the answer, the other asymptote should be $y=-\frac{1}{2}(2x+1)$.
How can I find it and how did I miss it in my calculations? Do I need to try and calculate the limit another way?
| You have the wrong value for $m_2.$
If $x \leq -1,$ then $\frac{\sqrt{x^2+x}}{x} \neq \sqrt{1+ \frac{1}{x}}.$
Do you see why?
Once you have fixed this error, the rest of your calculations should work OK.
| {
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"url": "https://math.stackexchange.com/questions/3046561",
"timestamp": "2023-03-29T00:00:00",
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Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \frac{-1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$ Task
Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = -\frac{1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \sum_{n=2}^{\infty} \frac{1}{3^n \cdot n (n-1)} - \frac{1}{3^n \cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ \sum_{n=1}^{\infty} \frac{1}{n\cdot p^n} = ln\frac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
| Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$\frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
$$\frac{1}{(n-1) n} = \frac{1}{n-1} - \frac{1}{n}$$
$$\frac{1}{(n-1)(n+1)} = \frac{1/2}{n-1} - \frac{1/2}{n+1}$$
$$\frac{1}{(n-1)n(n+1)} = \frac{1}{(n-1)n} - \frac{1}{(n-1)(n+1)} = \frac{1}{n-1} - \frac{1}{n} - \frac{1/2}{n-1} + \frac{1/2}{n+1} = \frac{1/2}{n-1} - \frac{1}{n} + \frac{1/2}{n+1}$$
$$\frac{2}{3^n (n-1) n (n+1)} = \frac{1}{3^n (n-1)} - \frac{2}{3^n n}
+ \frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$\begin{align}\sum_{n \ge 2} \frac{2}{3^n (n-1) n (n+1)} &= \frac{1}{3^2} + \left(\frac{1}{3^3} - \frac{2}{3^2}\right)\frac{1}{2} + \sum_{n \ge 3} \left(\frac{1}{3^{n-1}} - \frac{2}{3^{n}} + \frac{1}{3^{n+1}}\right) \frac{1}{n} \\ &= \frac{1}{54} + \left(3 - 2 + \frac{1}{3}\right)\sum_{n \ge 3} \frac{1}{3^n n} \\ &= \frac{1}{54} + \frac{4}{3} \sum_{n\ge 3} \frac{1}{3^n n} \\ &= \frac{1}{54} - \frac{4}{3} \left(\frac{1}{3} + \frac{1}{9 \cdot 2} \right) + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n} \\ &= - \frac{1}{2} + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n}. \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find $p, q, r$ prime numbers such that $r^2 - q^2 - p^2 = n^2$. Find $p, q, r$ prime numbers $p < q < r$ such that there exists natural number $n$ such that $r^2 - q^2 - p^2 = n^2$.
What I have done so far is obtaining the smallest number $p$ which is $2$.
The reason is that if we assume that $p$ is odd, as $p$ is the smallest of the 3 prime numbers, then $q, r$ are also odd and thus so is $n$.
Then, if we rewrite the initial relation as: $(r - p)(r + p) = n^2 + q^2$, we see that 4 divides $(r - p)(r + p)$, so 4 must also divide $n^2 + q^2$. But as $n$ and $q$ are odd, we have that $n^2 + q^2 = (2k + 1)^2 + (2l + 1)^2 = 4(k^2 + l^2 + k + l) + 2$ which is obviously not a multiple of 4.
Therefore, $p$ must be equal to 2.
For $q$ and $r$ I tried using other similar divisors arguments (as I believe the only solution is $(2,3,7)$), but none of them work, so if you have any suggestions, I would be very thankful.
| Assume $q$ and $r$ are primes bigger than $3$. Then they are not a multiple of $3$, so their squares are $1$ mod $3$. Then $r^2-q^2-p^2$ will be $2$ mod $3$, so not a square.
So we need $q=3$. Then we need $r^2-13=n^2$ and there are only finitely many pairs of squares differing by $13$, and as you found $7^2-13=6^2$ is the only solution.
| {
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"url": "https://math.stackexchange.com/questions/3048775",
"timestamp": "2023-03-29T00:00:00",
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Radius of convergence of $\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n$ I want to determine the convergence of the following series in dependency of $x$:
$\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n=\frac{3}{4}x+\frac{2}{5}x^2+\frac{1}{4}x^3+\frac{3}{17}x^4+ ... $
How can I solve this?
EDIT:
@Winther said, I should try the ratio test:
$q := \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n}\right| $
So we get
$q = \lim_{n \to \infty} \left| \frac{\frac{n+3}{2(n+1)^2+2}}{\frac{n+2}{2n^2+2}}\right| = \lim_{n \to \infty} \left| \frac{n+3}{2(n+1)^2+2}\frac{2n^2+2}{n+2} \right| = \lim_{n \to \infty} \left| \frac{2n^3+2n+6n^2+6}{2(n+1)^2n+4(n+1)^2+2n+4}\right| = \lim_{n \to \infty} \left| \frac{n^3+3n^2+n+3}{n^3+4n^2+6n+4}\right| $
With the tip from @Alex Vong to divide by $n^3$ the ratio test becomes:
$ q = \lim_{n \to \infty} \left| \frac{1+3/n+1/n^2+3/n^3}{1+4/n+6/n^2+4/n^3}\right|= 1$
So now there is no clear statement if the series is convergent ($q<1$ convergent, $q>1 $ divergent).
Should I try another test (e. g. the root test)?
EDIT 2: Corrected the first coefficients.
| Powers of $n$
do not affect the radius of convergence,
only convergence at the endpoints
(since
$(n^k)^{1/n}
\to 1$).
Therefore
$f(x)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} x^n
$
has the same radius of convergence as
$\sum\limits_{n=1}^{\infty} x^n$
which is
$-1 < x < -1$.
At $x=1$,
the series is
$f(1)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2}
$
which diverges like the
harmonic series.
At $x=-1$,
the series is
$f(-1)
=\sum\limits_{n=1}^{\infty} \frac{n+2}{2n^2+2} (-1)^n
$
which converges
because it is
an alternating series
with decreasing terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}{z}\right)\\
\frac{13}{z}=xyz\left(\frac{1}{y}+\frac{1}{x}\right)\end{cases}
Let $p=xyz.$ Then we have:
\begin{cases}\frac{1}{x}-\frac{p}{y}-\frac{p}{z}=0\\
\frac{p}{x}-\frac{8}{y}+\frac{p}{z}=0\\
\frac{p}{x}+\frac{p}{y}-\frac{13}{z}=0\end{cases}
Then we get
$$\frac{p+1}{x}=\frac{p+8}{y}=\frac{p+13}{z}$$
Any clue here?
| Let $a=1/x, b=1/y, c=1/z$.
$$a-pb-pc=0$$
$$pa-8b+pc=0$$
$$pa+pb-13c=0$$
From here, you can get $$(p+1)a=(13+p)c\\\text{and}$$
$$(p+1)a=(p+8)b$$
Substitute those values for $b$ and $c$ in last equation to get
$$p^3+11p^2-52=0$$
It'll give you p, and then you can find the corresponding values of $x,y$ and $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Rationalizing denominator of $\frac{18}{\sqrt{162}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{18}{\sqrt{162}}$$
The solution is provided:
$\sqrt{2}$
I arrived at:
$$\frac{\sqrt{162}}{9}$$
Here is my thought process to arrive at this incorrect answer:
$\frac{18}{\sqrt{162}}$
= $\frac{18}{\sqrt{162}}$ * $\frac{\sqrt{162}}{\sqrt{162}}$
= $\frac{18\sqrt{162}}{162}$
= $\frac{\sqrt{162}}{9}$
How can I arrive at $\sqrt{2}$ ?
| $$\require{cancel}\frac{18}{\sqrt{162}}=\frac{2\cdot3^2}{\sqrt{2\cdot3^4}}=\frac{2\cdot\cancel{3^2}}{\sqrt{2}\cdot\cancel{3^2}}=\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\cancel2\sqrt{2}}{\cancel{2}}=\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that a group with $3$ elements is cyclic?
Prove that a group with $3$ elements is cyclic.
I tried the case where $G=\{e,a,b\}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
Prove that a group with $4$ elements may or may not be cyclic.
| Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
\begin{array}{c|ccc}
& e & a & b \\ \hline
e & e & a & b \\
a & a & \\
b & b & \\
\end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
\begin{array}{c|ccc}
& e & a & b \\ \hline
e & e & a & b \\
a & a & b\\
b & b & \\
\end{array}
and now the diagram has a unique completion
\begin{array}{c|ccc}
& e & a & b \\ \hline
e & e & a & b \\
a & a & b & e \\
b & b & e & a\\
\end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
\begin{array}{c|cccc}
& e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & \\
b & b & \\
c & c &
\end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
\begin{array}{c|cccc}
& e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & e\\
b & b & \\
c & c &
\end{array}
Then we are forced to put in the following
\begin{array}{c|cccc}
& e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c\\
c & c & b
\end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
\begin{array}{c|cccc}
& e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e\\
c & c & b &
\end{array}
Now we can complete:
\begin{array}{c|cccc}
& e & a & b & c \\ \hline
e & e & a & b & c \\
a & a & e & c & b \\
b & b & c & e & a\\
c & c & b & a & e
\end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by $\{id,(12)(34),(13)(24),(14)(23)\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all $p \in \mathbb{Z}$ such that $ p^2+ 4p + 16 $ is a perfect square I've tried some things involving modular residues but they don't seem to work. Anyone know how to do this?
| So, we need $(p+2)^2+12=m^2$ for some integer $m$
$-12=(p+2-m)(p+2+m)$
As $p+2-m+p+2+m$ is even, the multiplicands have the same parity $\implies $ both must be ven
Consequently, $$\dfrac{p+2-m}2\cdot\dfrac{p+2+m}2=-3=1(-3)=(-1)3$$
Alternatively let $p^2+4p+16=(p+a)^2$ for some integer $a$
$\implies p=\dfrac{a^2-16}{4-2a}$
$\iff-2p=\dfrac{a^2-16}{a-2}=a+2-\dfrac{12}{2-a}$
$\iff12=(2-a)(a+2+2p)$
As $a+2+2p+2-a$ is even, $a+2+2p,2-a$ must have the same parity which must even as $12$ is even
$\implies\dfrac{2-a}2\cdot\dfrac{2+a+2p}2=3=(-3)(-1)=3\cdot1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^n \, dy = \frac{1}{n + 1},$$
the sum can be written as
$$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty H_n (xy)^n \, dx dy.\tag1$$
Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.
Making use of the generating function for the Harmonic number, namely
$$-\frac{\log(1-x)}{1 - x}~=~\sum_{n=1}^\infty H_n x^n,$$
the infinity sum appearing inside the integral in (1) can be written as
$$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = -\int_0^1 \int_0^1 \frac{\ln (1 - xy)}{x(1 - xy)} \, dx dy.$$
Enforcing a substitution of $x \mapsto x/y$ gives
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} &= -\int_0^1 \int_0^y \frac{\ln (1 - x)}{x (1 - x)} \, dx dy\\
&= -\int_0^1 \int_0^y \left [\frac{\ln (1 - x)}{x} + \frac{\ln (1 - x)}{1 - x} \right ] \, dx \, dy\\
&= \int_0^1 \operatorname{Li}_2 (y) \, dy + \frac{1}{2} \int_0^1 \ln^2 (1 - y) \, dy\\
&= I_1 + \frac{1}{2} I_2.
\end{align}
Here $\operatorname{Li}_2 (y)$ is the dilogarithm function.
For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have
\begin{align}
I_1 &= \int_0^1 \sum_{n = 1}^\infty \frac{y^n}{n^2} \, dy\\
&= \sum_{n = 1}^\infty \frac{1}{n^2} \int_0^1 y^n \, dy\\
&= \sum_{n = 1}^\infty \frac{1}{n^2(n + 1)}\\
&= \sum_{n = 1}^\infty \left [\frac{1}{n^2} + \frac{1}{n + 1} - \frac{1}{n} \right ]\\
&= \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \left (\frac{1}{n} - \frac{1}{n + 1} \right )\\
&= \frac{\pi^2}{6} - 1,
\end{align}
where the first of the series is the well-known Basel problem while the second telescopes.
For the second of the integrals, define
$$I(a) = \int_0^1 (1 - y)^a \, dy, \quad a > - 1.$$
Observe that
$$I''(0) = \int_0^1 \ln^2 ( 1 - y) \, dy,$$
where the derivative is with respect to $a$. Now, as
$$I(a) = \operatorname{B} (1, a + 1) = \frac{\Gamma (a + 1)}{\Gamma (a + 2)} = \frac{1}{a + 1},$$
where $\operatorname{B}(x,y)$ is the Beta function, one readily has
$$I''(a) = \frac{2}{(a + 1)^3},$$
yielding $I''(0) = I_2 = 2$.
Thus
$$\sum_{n = 1}^\infty \frac{H_n}{n (n + 1)} = \frac{\pi^2}{6} - 1 + \frac{1}{2} \cdot 2 = \frac{\pi^2}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example:
\begin{align*}
360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\
180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\
90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\
45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\
22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\
11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\
5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9
\end{align*}
As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!
| A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Find the locus of the foot of perpendicular from the centre of the ellipse. Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such that $P(a\cos\theta , b\sin\theta),\; Q(a\sin\theta, -b\cos\theta).\;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $\;(X_1,Y_1)\;$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.
| Let the centre of the ellipse be O and the foot be R. Also let point P is $(acos\theta, bsin\theta)$ then Q is $(acos(\theta+\frac{\pi}{2}), bsin(\theta+\frac{\pi}{2})$ or $(-asin\theta, bcos\theta)$ then PQ is
$$\frac{y - bcos\theta}{x + asin\theta} = \frac{b(sin\theta - cos\theta)}{a(sin\theta + cos\theta)} or \frac{\frac{y}{b} - cos\theta}{\frac{x}{a} + sin\theta} = \frac{sin\theta - cos\theta}{sin\theta + cos\theta}$$
Also OR is
$$\frac{y}{x} = - \frac{a(sin\theta + cos\theta)}{b(sin\theta - cos\theta)} or \frac{by}{ax} = -\frac{sin\theta + cos\theta}{sin\theta - cos\theta}$$
From these two equations we have
$$x^2 + y^2 = - axsin\theta + bycos\theta$$
Divide both sides by $\sqrt{a^2x^2 + b^2y^2}$ and let $tan\phi = \frac{by}{ax}$ then we have
$sin(\theta - \phi) = \frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}$ or $\theta = sin^{-1}(\frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}) + \phi$
Also from the second equation
$$\frac{y}{x} = - \frac{a}{b}\frac{tan\theta + 1}{-1 + tan\theta}$$
Substitute $\theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Expected area of triangle inscribed in a circle On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .
| For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2\phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = \sqrt{(1 - (\frac{2}{\pi})^2}$ and $sin\phi = \frac{2}{\pi}$
Let the line joining A and the centre makes an angle $\theta$ with the x axis, then
For $\theta$ between $0$ and $\frac{\pi}{2}$,
$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta + D)$$
For $\theta$ between $0$ and $\frac{\pi}{2} - \phi$
$$Area = \frac{1}{2}\frac{4}{\pi}(D - sin\theta)$$
For $\theta$ between $\frac{\pi}{2} - \phi$ and $\frac{\pi}{2}$
$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta - D)$$
Hence the average area is
$$\frac{4}{2\pi^2}(\int_0^{\frac{\pi}{2}}(sin\theta + D)d\theta + \int_0^{\frac{\pi}{2} - \phi}(D - sin\theta)d\theta + \int_{\frac{\pi}{2} - \phi}^{\frac{\pi}{2}}(sin\theta - D)d\theta)$$
$$= \frac{2}{\pi^2}(\frac{2}{\pi} + D\pi - 2D\phi + sin\phi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $\frac{3}{2\pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix demonstration $A^k$ Given a matrix $A = \begin{bmatrix} 7 & 4\\ -9 & -5 \end{bmatrix}$ $\in \mathcal{M2\times2}\, (\mathbb{R}) $
Show that $A^k = \begin{bmatrix} 1+6k & 4k\\ -9k & 1-6k \end{bmatrix} $
for every $k \in \mathbb{N}$
| Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = \begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 1 + 6 \cdot 1 & 4 \cdot 1 \\ -9 \cdot 1 & 1 - 6 \cdot 1 \end{bmatrix}; \tag 1$
then assuming that for some $k$
$A^k = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}, \tag 2$
we find
$A^{k + 1} = A^kA = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}\begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \\ -63k - 9 + 54k & -36k - 5 + 30k \end{bmatrix}$
$= \begin{bmatrix} 7 + 6k & 4 + 4k \\ -9 -9k & - 5 -6k \end{bmatrix} = \begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \\ -9(k + 1) & 1 - 6(k + 1) \end{bmatrix}, \tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k \ge 1$, as desired. $OE\Delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.
| I would use logarithmic differentiation. Setting $u(x)=x^2\cos x$, we have
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{u'(x)}{u(x)\ln u(x)}-\frac12\frac{-2x}{1-x^2} =\frac{2x\cos x-x^2\sin x}{x^2\cos x\,\ln(x^2\cos x)}-\frac12\frac{-2x}{1-x^2}\\[1ex]
&=\frac{2\cos x-x\sin x}{x\cos x\,\ln(x^2\cos x)}+\frac x{1-x^2},
\end{align}
whence
$$f'(x)=\frac{f'(x)}{f(x)}\,f(x)=\frac{2\cos x-x\sin x}{x\cos x\sqrt{1-x^2}}+\frac{x\ln(x^2\cos x)}{(1-x^2)^{3/2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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minimum value of $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
If $a,b,c>0.$ Then minimum value of
$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$
Try: Arithmetic geometric inequality
$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$
and $(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$
so $(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2\geq 18\sqrt{2}(abc)^{-1/3}$
could some help me to solve it. answer is $64$
| Hint: Another way is to consider Hölder's Inequality
$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 \geqslant (2+1+1)^3$$
Equality is possible when $8a^3=b^3=c^3=1$ (why?), so this is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Extremum value of $f(x,y)=ax^2+2hxy+by^2$ subject to constraints $g(x,y)=x^2+y^2-c^2=0$. Find the extremum value of $f(x,y)=ax^2+2hxy+by^2$ subject to constraints $g(x,y)=x^2+y^2-c^2=0$, where $abc \neq 0$ and $(a+1)^2+4(a^2-b^2) \geq 0$.
My attempt: I have use Lagrange Multipliers to solve this. Let us assume $$\phi(x,y, \lambda)=ax^2+2hxy+by^2+ \lambda (x^2+y^2-c^2).$$ Then $$\triangle _{x,y,\lambda}\phi =(2ax+2hy+2\lambda x,\> 2by+2hx+2 \lambda y,\> x^2+y^2-c^2).$$ Hence $$\triangle _{x,y,\lambda}\phi =0$$ implies \begin{align} (a+\lambda)x+hy & =0\\ hx+(b+\lambda)y &=0\\ x^2+y^2-c^2 & =0. \end{align} Then $$x=\frac{-(b+\lambda)y}{h} \implies x \neq 0, \> y \neq 0 \text{ as } abc \neq 0.$$ Now \begin{align} &(a+\lambda)x+hy =0\\ \implies & (a+ \lambda)(b+\lambda)y=h^2y \\ \implies & y=0 \text{ or } {\lambda}^2+(a+b)\lambda +(ab-h^2)=0. \end{align} Since $y \neq 0$ we have ${\lambda}^2+(a+b)\lambda +(ab-h^2)=0.$ Then \begin{align}\lambda & =\frac{-(a+b) \pm \sqrt{(a+b)^2-4ab+4h^2}}{2} \\ &= \frac{-(a+b) \pm \sqrt{(a-b)^2+4h^2}}{2}\end{align} But I cannot proceed further. PLease help me to find the extremum value of $f$.
| An alternative approach is using polar coordinates. WLOG, let $c=1$. Then you need to find the extrema of
$$a\cos^2t+2h\cos t\sin t+b\sin^2t=\frac{a-b}2\cos2t+\frac{a+b}2+h\sin2t.$$
As this expression describes a sinusoid, the values of the extrema are immediate:
$$\frac{a+b\pm\sqrt{(a-b)^2+4h^2}}2.$$
(If $c\ne1$, times $c^2$.)
| {
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Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I proceed to induct on $m$ (and on $m$ only).
I am not perfectly confident in my self, however, for I see two placeholders, $n$ and $m.$ Is this a case where double induction is needed (first on $m$ and then on $n$)?
Whether or not this proof requires double induction, may someone explain when double induction is needed?
Consider any fixed $n, r \geq 0$ and the following two cases (I know that only one case is needed to complete this inductive proof).
CASE 1
\begin{align}
\binom{n + 0}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{0}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r}\binom{0}{0} \\ &= 0 + 0 + \cdots + \binom{n}{r} \\ &= \binom{n}{r}
\end{align}
CASE 2
\begin{align}
\binom{n + 1}{r} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{1}{r - i} \\ &= \binom{n}{0}\binom{0}{r} + \binom{n}{1}\binom{0}{r-1} + \cdots + \binom{n}{r-1}\binom{1}{r - (r-1)} + \binom{n}{r}\binom{1}{r - r} \\ &= 0 + 0 + \cdots + \binom{n}{r-1} + \binom{n}{r} \\ &= \binom{n}{r-1} + \binom{n}{r}
\end{align}
INDUCTION
Suppose it is true for $m \leq k.$ Now, consider $$\binom{n + (k + 1)}{r}.$$ It follows from Pascal's Identity that
$$\binom{n + (k+1)}{r} = \binom{n + k}{r} + \binom{n + k}{r-1}$$
And,
\begin{align}
\binom{n + k}{r} + \binom{n + k}{r-1} &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - i} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k}{r - 1 - i} \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\bigg[\binom{k}{r - i} + \binom{k}{r - 1 - i}\bigg] \\ &= \binom{n}{r} + \sum_{i = 0}^{r-1} \binom{n}{i}\binom{k+1}{r-i} \\ &= \sum_{i = 0}^{r} \binom{n}{i}\binom{k+1}{r-i}
\end{align}
Hence, the equality holds for $m = k + 1.$ Given that the equality holds for $m = 0, 1,$ and that if equality holds for $m = k,$ it then holds for $m = k + 1,$ it follows that the equality holds $\forall m \in \mathbb{N}.$
| Note that$$(x+1)^{n+m}=(x+1)^n(x+1)^m$$
Then by binomial theorem and collecting terms
\begin{align}
\sum_{r=0}^{n+m}\binom{n+m}{r}x^r &=
\sum_{i=0}^{n}\binom{n}{i}x^i\sum_{j=0}^{m}\binom{m}{j}x^j \\&=
\sum_{r=0}^{n+m}\sum_{i+j=r}\binom{n}{i}\binom{m}{j}x^r \\ &=
\sum_{r=0}^{n+m}\sum_{i=0}^r\binom{n}{i}\binom{m}{r-i}x^r
\end{align}
Then compare the coefficient
| {
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} |
Compute $ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$
Compute $$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$$
I did the following:
$$ \lim\limits_{n \to \infty}\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}} = \frac{\sqrt{3}}{\sqrt{1}}=\sqrt3$$
However, the correct answer is different. Why am I wrong?
Thank you for your help.
| By dividing the numerator and the denominator by $n$, you should have
$$\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}}=\frac{\sqrt{3+\frac{1}{n}-\frac{1}{n^2}}}{1+\sqrt{1-\frac{1}{n^2}}}.$$
Instead at your denominator we have $\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}$.
Then what is the final limit?
| {
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When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.
When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately appear, as a function of $x$?
It seems to be around ~ $c \sqrt x$.
Example: $x = 2500$:
2500, 1250, 833, 625, 500, 416, 357, 312, 277, 250, 227, 208, 192, 178, 166, 156, 147, 138, 131, 125, 119, 113, 108, 104, 100, 96, 92, 89, 86, 83, 80, 78, 75, 73, 71, 69, 67, 65, 64, 62, 60, 59, 58, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 43, 42, 41, 40, 40, 39, 39, 38, 37, 37, 36, 36, 35, 35, ...
| It cannot occur between term $n$ and term $n+1$ if $\frac{x}{n} - \frac{x}{n+1} \ge 1$, equivalently $x \ge n^2 + n$, equivalently $n \le -\frac{1}{2} + \frac{\sqrt{1+4x}}{2}$.
It must occur, either between term $n$ and $n+1$, or between term $n+1$ and $n+2$, if $\frac{x}{n} - \frac{x}{n+1} \le \frac{1}{2}$, equivalently $x \le \frac{1}{2} n^2 + \frac{1}{2} n$, equivalently $n \ge -\frac{1}{2} + \frac{\sqrt{1+8x}}{2}$.
So the first place it appears is somewhere between the two extremes of $-\frac{1}{2} + \frac{\sqrt{1+4x}}{2}$ and $-\frac{1}{2} + \frac{\sqrt{1+8x}}{2} + 1 = \frac{1}{2} + \frac{\sqrt{1+8x}}{2}$.
| {
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Why does this math trick work? 35 by 11 is 385 because 3+5 is 8, so it's the digit in the middle.
Same for:
72 by 11 is 792 because 7+2 is 9, so it's the digit in the middle.
I see it works because 35 by 10 is 350, or 72 by 10 is 720. The 0 is replaced with the extra digit. The last digit is 5 by 1 or 2 by 1, so it stays the same.
But why should the middle digit be the sum of the first and last?
| Let "$ab$" be a two digit number.
Then $ab = 10a + b$ and $ab\times 11 = (10a+b)(10 + 1) =$
$ 10a(10 + 1) + b(10+1) =$
$ (100a + 10a)+ (10b + b) =$
$100a + (10a + 10b)+b =$
$100a + 10(a+b) + b$.
And if $a+b < 10$ we get $100a + 10(a+b) + b = a(a+b)b$.
Not it doesn't work if $a+b \ge 10$. Example $84\times 11= 924$ and $9+4 =13$ and not $2$. But notice that $9+4 -11 = 2$.
If $a+b \ge 10$ you get:
$100a + 10(a+b) + b = 100a + 10([a+b-10] + 10)+b$
$=100a + 100 + 10[a+b-10] + b = (a+1)(a+b-10)b$.
If we write this as $cde$ we have $c+e = a+b + 1 = (a+b-10) + 11 = d+11$.
So you can modify to rule to if $cde = K\times 11$ then either $c +e =d$ or $c+e = d+11$
We can extend this further:
if $M$ is a multiple of eleven then if you add the even position digits together and add the odd position digits together the sums are equal or off by a multiple of $11$.
| {
"language": "en",
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Prove that $\sigma_n - e$ is decreasing faster than $e-S_n$ given specific $\sigma_n$ and $S_n$.
Let $\sigma_n$ and $S_n$ be defined as:
$$
\sigma_n = 3 - \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!} \\
S_n = 1 + \sum_{k=1}^n\frac{1}{k!}
$$
Show that $\sigma_n - e$ is decreasing faster than $e-S_n$.
I may use anything before the definition of a derivative.
Based on the question we want to eventually show that:
$$
\sigma_n - e \le e - S_n
$$
Lets adjust $\sigma_n$. Define $a_n$:
$$
a_n = \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!}
$$
This may be expanded by partial fractions:
$$
\begin{align}
a_n &= \sum_{k=1}^n\frac{1}{k(k+1)(k+1)!} \\
&= \sum_{k=1}^n\frac{1}{k(k+1)}\cdot\frac{1}{(k+1)!} \\
&= \sum_{k=1}^n\left(\frac{1}{k} - \frac{1}{k+1}\right)\cdot\frac{1}{(k+1)!} \\
&= \sum_{k=1}^n\left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2}\right)\cdot\frac{1}{k!} \\
&= \sum_{k=1}^n \frac{1}{kk!} - \color{red}{\sum_{k=1}^n \frac{1}{(k+1)k!}} - \color{blue}{\sum_{k=1}^n \frac{1}{(k+1)^2k!}} \\
&= \sum_{k=1}^n \frac{1}{kk!} - \color{blue}{\sum_{k=1}^n \frac{1}{(k+1)(k+1)!}} - \color{red}{\sum_{k=1}^n \frac{1}{(k+1)k!}} \\
\text{(telescoping)} &= 1 - \frac{1}{(n+1)(n+1)!} - \sum_{k=1}^n \frac{1}{(k+1)!}
\end{align}
$$
Now going back to the inequality:
$$
\sigma_n - e \le e - S_n \iff \\
\sigma_n + S_n \le 2e \iff \\
3 - a_n + S_n \le 2e
$$
Replacing the terms with actual sums one may obtain:
$$
\begin{align*}
3 - a_n + S_n &= 3 - \left(1 - \frac{1}{(n+1)(n+1)!} - \sum_{k=1}^n \frac{1}{(k+1)!}\right) + 1 + \sum_{k=1}^n{1\over k!} \tag{1.1} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=1}^n \frac{1}{(k+1)!} + \sum_{k=1}^n{1\over k!} \tag{1.2} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=2}^{n+1} \frac{1}{k!} + \sum_{k=1}^n{1\over k!} \tag{1.3} \\
&= 3 + \frac{1}{(n+1)(n+1)!} + \sum_{k=1}^{n} \frac{1}{k!} - 1 + {1\over (n+1)!} + \sum_{k=1}^n{1\over k!} \tag{1.4} \\
&= 2 + \frac{1}{(n+1)(n+1)!} + \frac{1}{(n+1)!} + 2\sum_{k=1}^{n} \frac{1}{k!} \tag{1.5} \\
&= 2\left(1 + \frac{1}{2(n+1)!}\left({1\over n+1} + 1\right) + \sum_{k=1}^{n} \frac{1}{k!}\right) \tag{1.6} \\
&= 2\left(\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!}\right) \tag{1.7}
\end{align*}
$$
Which eventually results into:
$$
\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!} \le e
$$
Now based on this question:
$$
e - S_n \le \frac{n+2}{(n+1)(n+1)!}
$$
While I wanted to show:
$$
\frac{n+2}{\color{red}{2}(n+1)(n+1)!} + S_n \le e \iff \\
\frac{n+2}{\color{red}{2}(n+1)(n+1)!} \le e - S_n
$$
Or summarizing:
$$
\frac{n+2}{2(n+1)(n+1)!} \le e - S_n \le \frac{n+2}{(n+1)(n+1)!}
$$
Which seems to be the case. Now I got stuck. How do I proceed from here?
Description of the steps:
*
*$(1.1)$ - replace $\sigma_n$ and $S_n$ with sums
*$(1.2)$ - cancel $-1+1$
*$(1.3)$ - change indexing in the sum
*$(1.4)$ - add and subtract $1$ and change the index of the sum. Fetch last term of the sum.
*$(1.5)$ - add the sums.
*$(1.6)$ - factor out $2$. Factor out $1\over (n+1)!$
*$(1.7)$ - inject $1$ into the sum and change indexing. Cast brackets to a single fraction
| Presuming my cursory review of your work leading up to the inequality $\frac{(n+2)}{2(n+1)(n+1)!} + \sum_{k=0}^{n} \frac{1}{k!} \le e$ didn't miss an error and assuming $n \in \mathbb N$, we may proceed as follows:
\begin{aligned}\\
e-\sum_{k=0}^{n} \frac{1}{k!}
&= \sum_{k=n+1}^{\infty} {1\over k!}\\
&\geq \frac1{(n+1)!}+\frac1{(n+2)!}\\
&\geq \frac1{2(n+1)!}+\frac1{(2n+2)(n+1)!} && (n+2\leq2n+2) \\
&=\frac{n+1}{2(n+1)(n+1)!}+\frac1{2(n+1)(n+1)!} \\
&=\frac{n+2}{2(n+1)(n+1)!}
\end{aligned}
Seems to me that you did most of the work in your original question.
| {
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Complex analysis proof triangle inequality: Given: $|z+w|^2=|z|^2+|w|^2+2Re(z\bar w)$
Prove:$|z+w|\leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=\sqrt{(x+a)^2+(y+b)^2}$$
$$\sqrt{x^2+y^2}+\sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.
Any help is appreciated!
| You want to show that
$$\tag1
\sqrt{(x+a)^2+(y+b)^2}\leq \sqrt{a^2+b^2}+\sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$\tag2
{(x+a)^2+(y+b)^2}\leq a^2+b^2+x^2+y^2+2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2},
$$
which reduces to
$$\tag3
2ax+2yb\leq 2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$\tag4
0\leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2\leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2by\leq|2ax+2by|\leq2\sqrt{a^2+b^2}\,\sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
| {
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Making $x$ the subject of $x^3-2x+y=1$ How to I make x the subject of this formula?
$$x^3-2x+y=1\tag1$$
My attempt:
I apply complete the square
$$(x-1)^3+y-1=1\tag2$$
$$(x-1)^3=2-y\tag3$$
$$x-1=\sqrt[3]{2-y}\tag4$$
$$x=1+\sqrt[3]{2-y}\tag5$$
but my teacher said it is wrong!
Can you please help? Thank in advance!
| Use the formula for a cubic and find three solutions for $x$:
*
*$\frac{\sqrt[3]{2} \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}+4 \sqrt[3]{3}}{6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$
*$\frac{i \sqrt[3]{2} \left(\sqrt{3}+i\right) \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}-4 \sqrt[3]{3} \left(1+i \sqrt{3}\right)}{2\
6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$
*$\frac{4 i \sqrt[3]{3} \left(\sqrt{3}+i\right)-\sqrt[3]{2} \left(1+i \sqrt{3}\right) \left(\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9\right)^{2/3}}{2\
6^{2/3} \sqrt[3]{\sqrt{3} \sqrt{27 y^2-54 y-5}-9 y+9}}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Probability of extracting a ball after two balls were swapped We have $2$ boxes, the first one contains $10$ white balls and $11$ black balls. The second box contain $12$ white balls and $13$ black balls.
We swap two balls between the boxes then we extract a ball from the first box. What is the probability that the ball is white?
A swap consists in taking a ball from the first box and put it in the second one, then taking a ball from the second box and put it in the first one.
$$P(\text{white ball from box1})=\frac{10}{21}$$
$$P(\text{black ball from box1})=\frac{11}{21}$$
$$P(\text{white ball from box2})= \frac{12+1}{26}\frac{10}{21}+\frac{12}{26}\frac{11}{21}=\frac{13\cdot10+12\cdot11}{26\cdot 21}$$
$$P(\text{black ball from box2})= \frac{13}{26}\frac{10}{21}+\frac{13+1}{26}\frac{11}{21}=\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$
$$P(\text{white ball from box1 after swap})=\frac{10+1}{22}\cdot\frac{13\cdot10+12\cdot11}{26\cdot 21}+\frac{10}{22}\cdot\frac{13\cdot 10+14\cdot 11}{26\cdot 21}$$
$$=\frac{2882}{12012}+\frac{2840}{12012}=\frac{5722}{12012}=0.4763$$
Have I done it correctly? I don't know how to verify myself with this kind of problems.
Also, is there a trick to find the probability if there were two swaps?
| The probability of ending up with an additional white ball equals:
$$\frac{11}{21} \frac{12}{26}$$
The probability of ending up with an additional black ball equals:
$$\frac{10}{21} \frac{13}{26}$$
We thus find an overall probability of:
$$\frac{11}{21} \frac{12}{26} \frac{11}{21} + \frac{10}{21} \frac{13}{26} \frac{9}{21} + \left(1 - \frac{11 \cdot 12 + 10 \cdot 13}{21 \cdot 26}\right) \frac{10}{21} = \frac{1452 + 1170 + 2840}{11466} = \frac{5462}{11466} \approx 0.4764$$
| {
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Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.
| Well, if you hit a snag that means either... the thing you are trying to prove is false... or it's not actually a snag.
You say $6^{n+5}\equiv 76*6^n \pmod {100}$ does not give the correct result.
How do you know $76*6^n\not \equiv 6^n \pmod {100}$?
...
$76*6^n \equiv 6^n \pmod {100} \iff$
$75*6^n \equiv 0 \pmod {100} \iff$
$100 |75*6^n \iff$
$4|3*6^n = 3^{n+1}2^n \iff$ (as $2$ is prime)
$4|2^n \iff$
$n \ge 2$.
So your snag is only a snag if $n < 2$ in which case the result isn't true.
....
Alternatively we can use Chinese Remainder Theorem.
If $76*6^n \equiv k \pmod{100}$ then there is a unqiue solution to
$k \equiv 76*6^n \equiv 6^n \pmod{25}$ and to
$k \equiv 0 \pmod 4$
If $n = 0$ then $6^n \equiv 1 \pmod 4$ and our solution is $k = 6^n + 25m$ where $25m\equiv m \equiv 3\pmod 4$. (i.e. $m = 3$ and $6^5 = 7776 \equiv 76 = 1 + 75 = 6^0 + 75 \pmod {100}$).
If $n = 1$ then $6^n \equiv 2\pmod 4$ and our solution is $k = 6^n + 25 m $ where $25m \equiv m \equiv 2\pmod 4$ (i.e. $m = 2$ and $6^6= 46656 \equiv 56\equiv 6^1 + 50 \pmod {100}$).
If $n \ge 2$ then $6^n \equiv 0 \pmod 4$ and ore solution is $k =6^n + 25m$ where $25m \equiv m \equiv 0 \pmod 4$ and $6^{n+5}=7776*6^n \equiv 76*6^n = 75*6^n + 6^n \equiv 6^n \pmod {100}$.
...
Third option. $76*6^n = 50*6^n + 25*6^n + 6n^2 = 100(\frac {6^n}2+ \frac {6^n}4) + 6^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$ solve $$\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt$$
my attempt
$$t = \frac{1}{2}\tan(u)$$
$$dt = \frac{1}{2}\sec^2(u)du\\$$
$$\begin{align}
\int_{0}^{1} t^2 \sqrt{(1+4t^2)}dt&=\int_{0}^{1} \frac{\tan^2(u)}{4} \sqrt{1+\tan^2(u)}\frac{1}{2}\sec^2(u)du\\
&=\frac{1}{8}\int_{0}^{1} \tan^2(u)\sec^{3}(u)du\\
&= \frac{1}{8} \int_{0}^{1} (\sec^2(u) - 1)(\sec^{3}(u))du\\
&=\frac{1}{8}\int_{0}^{1} \sec^5(u)du - \frac{1}{8} \int_{0}^{1}\sec^3(u)du
\end{align}$$
what now?
| $$\begin{align}
\int t^2 \sqrt{(1+4t^2)}dt&=\frac{1}{8}\int{\sinh^2(x)}\sqrt{1+\sinh^2(x)}\cosh(x)dx\\
&=\frac{1}{8}\int{\sinh^2(x)}\cosh^2(x)dx=\frac{1}{8\times 32}\left(\sinh(4x)-4x\right) +C \\
&= \frac{1}{256}\left((4\sinh(x)\cosh^3(x) + 4\sinh^3(x)\cosh(x)) - 4x\right) + C\\
&=\left[\frac{1}{256}\left((4(2t)(\sqrt{1+4t^2})^3 + 4(2t)^3\sqrt{1+4t^2}) - 4\sinh^{-1}(2t)\right)\right]_{0}^{1} \approx 0.6063
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095152",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simplifying $\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$ In simplifying
$$\frac{2x^2-5x-3}{6x^3-2x^4}$$
I got this far
$$\frac{ (2x+1) (x-3) }{ 2x^3 (3-x) }$$
but there aren't same brackets to cancel out.
| $\frac{(2x+1)(x-3)}{2x^3(3-x)}= -\frac{(2x+1)(x-3)}{2x^3(x-3)}= -\frac{(2x+1)}{2x^3} $
| {
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"answer_count": 3,
"answer_id": 2
} |
Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$
I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
| Let $\arctan x=y\implies x=\tan y,x^2-1=\dfrac{\sin^2y}{\cos^2y}-1=-(1+x^2)\cos2y$
$$2x\cos(2x)+(x^2-1)\sin2x=(1+x^2)\sin2(\arctan x-x)$$
$$I=\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} =\int\dfrac{2x^2}{(1+x^2)\sin2(\arctan x-x)}$$
Now set $u=\arctan x-x,du=-\dfrac{x^2\ dx}{1+x^2}$
$$I=-2\int\dfrac{du}{\sin2u}=?$$
Use this
| {
"language": "en",
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"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$
Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$
$$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$
How can I solve it?
| This is a variant of TheSimpliFire's approach given in a comment.
By letting $x=e^t$ we get
$$\begin{align*}
\int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}\,dx
&=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\
&=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,dt\\
&=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+1)t}\,dt+\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\
&=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+1)^3}+\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}\\
&=\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)+\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}-\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)\\
&=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=2\left(\sum_{n=0}^\infty\frac{1}{n^2}-\sum_{n=0}^\infty\frac{1}{(2n)^2}\right)\\&=2\left(1-\frac{1}{4}\right)\sum_{n=0}^\infty\frac{1}{n^2}=\frac{3}{2}\cdot \frac{\pi^2}{6}=\frac{\pi^2}{4}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$ Show $\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
from LS
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(3/2) + \log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2\big(\frac{6(n+1)}{6}\big)}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
is this right?
| Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.
1)$$
\log_{a}{\frac{x}{y}}=\log_{a}{\left[\left(\frac{y}{x}\right)^{-1}\right]}=
-1\cdot\log_{a}{\frac{y}{x}}=-\log_{a}{\frac{y}{x}}.
$$
2)$$
\log_{a}\left({xy}\right)=\log_{a}{x}+\log_{y}.
$$
$$
1+\frac{\log_2\left[\frac{2}{3}(n+1)\right]}{\log_2\frac{3}{2}} =
1+\frac{\log_2\frac{2}{3}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}} =\\
1+\frac{-\log_2\frac{3}{2}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=
1-\frac{\log_2\frac{3}{2}}{\log_2\frac{3}{2}}+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=\\
1-1+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=
\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding weight in Megagrams (Mg) given circumference and density Glaciers often deposit large rocks called erratics. The granite rock has a circumference of 9.5 m. Assuming it conforms to the shape of a sphere, what would be its weight in Megagrams (Mg), where 1 Mg = 1,000 Kg ≈ 1 US ton. The average density of granite is $2.70\ \mathrm{g} \cdot \mathrm{cm}^{-3}$.
| HINT
*
*Given a circumference of the sphere, find its volume $V$.
*With volume and density, find the mass.
*Convert the mass to the desired units.
UPDATE
Since the circumference is $$C=2\pi r \iff r = \frac{C}{2\pi}$$ and $V = 4\pi r^3/2$, you have
$$
\begin{split}
V &= \frac43\pi r^3
= \frac43 \pi \left(\frac{C}{2\pi}\right)^3
= \frac{4}{3 \cdot 8} \frac{\pi}{\pi^3} C^3
= \frac{C^3}{6\pi^2} \\
&= \frac{(9.5 \mathrm{m})^3}{6 \pi^2}
= \frac{9.5^3}{6 \pi^2} \textrm{m}^3 \\
&\approx 14.478 \mathrm{m}^3.
\end{split}
$$
Your density $\rho$ is in the wrong units, so to find the mass $M$ you have to do the following:
$$
\begin{split}
M &= \rho V
= 2.7 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 14.478 \mathrm{m}^3 \\
&= 2.7 \frac{\mathrm{g}}
{\mathrm{\left(cm \times \frac{1 \mathrm{m}}
{100 \mathrm{cm}}\right)}^3}
\times 14.478 \mathrm{m}^3 \\
&= \frac{2.7 \mathrm{g}}{\left(\frac{1}{100} \mathrm{m}\right)^3}
\times 14.478 \mathrm{m}^3 \\ \\
&= 2.7 \times 100^3 \times 14.478 \frac{\mathrm{g} \cdot \mathrm{m}^3}{\mathrm{m}^3} \\
&= 39090.6 \mathrm{g}.
\end{split}
$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Proving $\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$
I would like to prove using Mean Value theorem for $n \ge 1$
$$\frac{n}{2n+1} < \sqrt{n^2+n} -n < \frac{1}{2}$$
RHS can be proved by rationalizing the square root term, not sure about the LHS.
| $n^2+n=(n+\frac12)^2-\frac14<(n+\frac12)^2\Rightarrow \sqrt{n^2+n}-n<\frac12$
$\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}>\frac{n}{n+\frac12+n}>\frac{n}{2n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Problem solving a 4 by 4 linear system with infinite solutions I have a question here. The following is a 4 by 4 system of linear equations which has infinitely many solutions.
$$
\left\{
\begin{array}{c}
x+y+3z+t=0 \\
x-y-z-t=0 \\
3x+y+5z+3t=0 \\
x+5y+11z+8t=0
\end{array}
\right.
$$
An online calculator gave me these solutions:
$t = 0, x = - r_1, y = -2 r_1, z = r_1$, where $r_1$ is a parameter.
However, how did they solve this? It's kinda complicated for me, because what I'd do is I'd isolate "$x$" from the second equation $(x=y+z+t)$, then I would plug that into the first equation and so on...
Can someone help? Thank you and sorry for taking your time !
| Use the matrix of the linear system and put it in reduced row echelon form:
\begin{align}
&\left[\begin{array}{rrrr}
1&1&3&1\\1&-1&-1&-1\\3&1&5&3\\1&5&11&8
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&1&3&1\\0&-2&-4&-2\\0&-2&-4&0\\0&4&8&7
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&1&3&1\\0&-2&-4&-2\\0&0&0&2\\0&0&0&3
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&1&3&1\\0&1&2&1\\0&0&0&1\\0&0&0&3
\end{array}\right] \\[1ex]
&{}\rightsquigarrow\left[\begin{array}{rrrr}
1&1&3&0\\0&1&2&0\\0&0&0&1\\0&0&0&0
\end{array}\right]\rightsquigarrow
\left[\begin{array}{rrrr}
1&0&1&0\\0&1&2&0\\0&0&0&1\\0&0&0&0
\end{array}\right].
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103678",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$? I need to find the limit of $\frac {x^2+x} {x^2-x-2}$ where $x \to -1$. Right now I am getting $\frac{0}{0}$ if I don't factor first, or $\frac{2}{0}$ if I do.
Here are my factoring steps:
$\frac {x^2+x} {x^2-x-2}$
$=\frac{x(x+1)}{(x-2)(x+1)}$
replace $x$ with $-1$
$=\frac{-1(-1+1)}{(-1-2)(-1+1)}$
$=\frac{2}{-3 (0)}$
$=\frac{2}{0}$
How can I solve this problem?
| Just factorise:
$$\frac{x^2+x}{x^2-x-2} \equiv \frac{x(x+1)}{(x-2)(x+1)}$$
Assuming $x \neq -1$, we can divide the numerator and denominator by $x+1$. For all $x \neq -1$, we have
$$\frac{x^2+x}{x^2-x-2} = \frac{x}{x-2}$$
As $x \to -1$, we see that
$$\frac{x^2+x}{x^2-x-2} \to \frac{-1}{-1-2} = \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$
Rewriting the expression
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{1}{a-2}-\dfrac{1}{a(a+2)}$$
$$\dfrac{2a}{(a-2)(a+2)} - \dfrac{(a+2)}{(a-2)(a+2)}-\dfrac{1}{a(a+2)}$$
Factoring $\dfrac{1}{a+2}$
$$\dfrac{1}{a+2}\biggr (\dfrac{2a}{a-2} - \dfrac{a+2}{a-2}-\dfrac{1}{a}\biggr )$$
This is where I'm stuck. Could you assist me?
Regards
| \begin{align}
\dfrac{1}{a+2}\biggr(\dfrac{2a}{a-2} - \dfrac{a+2}{a-2}-\dfrac{1}{a}\biggr)&=\dfrac{1}{a+2}\biggr (\dfrac{2a-a-2}{a-2}-\dfrac{1}{a}\biggr)\\
&=\dfrac{1}{a+2}\biggr(\dfrac{a-2}{a-2}-\dfrac{1}{a}\biggr )\\
&=\dfrac{1}{a+2}\biggr(1-\dfrac{1}{a}\biggr)\\
&=\dfrac{1}{a+2}\biggr(\dfrac{a-1}{a}\biggr)\\
&=\dfrac{a-1}{a (a+2)}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Functional equation $ (n+1) f(n+1)= (a n+b) f(n) $ for $n=0,1,...$ I am looking for a solution to the following functional equation:
\begin{align}
(n+1) f(n+1)= (a n+b) f(n), n=0,1,...
\end{align}
where $a$ and $b$ are some positive constants.
Moreover, $f(n)$ is positive and
\begin{align}
0<f(n) &\le 1,\\
\sum_{n=0}^\infty f(n)&=1.
\end{align}
I was able to find a solution if $a=b \le 1$
\begin{align}
f(n)= \frac{c}{(1+c)^{n+1}}, \, c=\frac{1-a}{a}.
\end{align}
My questions:
*
*Is there a systematic way to solve this equation?
*Is the solution to this equation unique?
Any reference would also be appreciated.
| One way is to use generating functions. Define
\begin{align*}
g(x) = \sum_{n=0}^{\infty} f(n) x^n
\end{align*}
Henceforth, I will use the notation
\begin{align*}
g(x) \leftrightarrow\{f(n)\}
\end{align*}
to indicate that $g(x)$ has coefficients of $f(n)$ for $x^n$ in its series expansion. Then we have
\begin{align*}
g'(x) \leftrightarrow\{(n+1)f(n+1)\}
\end{align*}
and
\begin{align*}
ax g'(x) + bg(x) \leftrightarrow \{(an+b)f(n)\}
\end{align*}
We can set equal the corresponding generating functions
\begin{align*}
g'(x) = axg'(x) + bg(x)
\end{align*}
which has solution
\begin{align*}
g(x) = c_1 (1 - ax)^{-b/a} = \sum_{n=0}^{\infty} c_1 (-a)^n \binom{-b/a}{n}x^n \leftrightarrow \left\{c_1 (-a)^n \binom{-b/a}{n}\right\}
\end{align*}
So,
\begin{align*}
f(n) = c_1 (-a)^n \binom{-b/a}{n}
\end{align*}
The constant $c_1$ can be found be evaluating
\begin{align*}
1 = \sum_{n=0}^{\infty} f(n) = g(1) = c_1 (1 - a)^{-b/a} \implies c_1 = (1 - a)^{b/a}
\end{align*}
So we find
\begin{align*}
f(n) = (1 - a)^{b/a} (-a)^n \binom{-b/a}{n}
\end{align*}
It's worth noting that $f(n)$ is always positive, so we also have $0 < f(n) \le 1$.
| {
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"source": "stackexchange",
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What is the area of $\triangle ABC$ where $\triangle ADC$ is cyclic, point $P$ is on the circumference and $AD = AP$?
$\triangle ABC$ is a right angled triangle. The perpendicular drawn form $A$ on $BC$ intersects $BC$ at point $D$. A point $P$ is chosen on the circle drawn through the vertices of $\triangle ADC$ such that $CP$ $\perp$ $BC$ and $AP$ = $AD$.A square is drawn on the side $BP$ and its area is 350 unit$^\text{2}$. What is the area of $\triangle ABC$?
My Attempt:
Here, $ADCP$ is a square because $AD = AP$ and $\angle ADC$ = $\angle DCP$ = 90$^\circ$. So, by expressing $AD = x$ and $BD$ = $y$,
From the right angled $\triangle ADB$,
$AB^\text{2}$ = $x^\text{2} + y^\text{2}$
And now, from the $\triangle ABC$,
$x^\text{2} + y^\text{2}$ + ($\sqrt 2 x$)$^\text{2}$ = $(x + y)$$^\text{2}$.....($AC$ is the diagonal of the square $ADCP$)
$x^\text{2}$ + $y^\text{2}$ + 2x$^\text{2}$ = $x^\text{2}$ + $y^\text{2}$ + 2$xy$ $\implies$ $x$ =$y$
So, $BD$ = $x$ and $AB$ = $\sqrt(2x^\text{2})$ $\implies$ $AB$ = $\sqrt2 x$. And now from $\triangle BCP$,
($2x$)$^\text{2}$ + $x$$^\text{2}$ = ($\sqrt350$)$^\text{2}$
$5x$$^\text{2}$ = $350$ $\implies$ $x$$^\text{2}$ = $70$ $\implies$ $x$ = $\sqrt70$
After that, the area of $\triangle ABC$ = $\frac{1}{2}$×$\sqrt2x$×$\sqrt2x$
= $\frac {1}{2}$×$2x$ $^\text{2}$ = $x^\text{2}$ = ($\sqrt70$)$^\text{2}$ = $70$.
Is my answer correct? If not so, can anyone please provide me with another solution or method for better learning process? Or simply any kind of clue or hint will be so much helpful. Thanks in advance.
| Since $ADCP$ is square, we obtain $AD=BD=DC=x$
because $\measuredangle ACD=45^{\circ}$ and from here also $\measuredangle ABD=45^{\circ}.$
Thus, $BC=2x$ and $$350=PB^2=(2x)^2+x^2,$$ which gives $x^2=70$ and
$$S_{\Delta ABC}=\frac{2x\cdot x}{2}=70.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that |z|=1. I am trying to prove that
If $z\in \mathbb{C}-\mathbb{R}$ such that $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$. Show that $|z|=1$.
1 method , through which I approached this problem is to assume $z=a+ib$ and to see that $$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$$.
So problem reduces to show that $|z|=1$ whenever $\frac{2z}{z^2-z+1}\in \mathbb{R}$
I put $z=a+ib$ and then rationalise to get the imaginary part of $\frac{2z}{z^2-z+1}$ be $\frac{b-b^3-a^2b}{something}$. I equated this to zero and got my answer.
Is there any better method?
| We are given that the imaginary part of $\frac{z^2+z+1}{z^2-z+1}$ is zero.
Therefore $\frac{z^2+z+1}{z^2-z+1}$ is equal to its own conjugate:
$$\frac{z^2+z+1}{z^2-z+1} = \frac{\bar z^2+\bar z+1}{\bar z^2-\bar z+1}.$$
Cross-multiply (multiply both sides by $(z^2-z+1)(\bar z^2-\bar z+1)$)
and cancel all terms on the right. The result is
$$ 2z\bar z^2 - 2z^2\bar z - 2\bar z + 2z = 0,$$
which you can simplify to
$$(\bar z - z)z\bar z - \bar z + z = 0.$$
Since the imaginary part of $z$ is not zero,
it follows that $\bar z - z \neq 0$ and you can divide both sides of the equation
by $\bar z - z$ to obtain
$$ z\bar z - 1 = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim\limits_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$ does not exist.
Show that $$\lim_{(x,y)\to(0,0)}\frac{x^3y-xy^3}{x^4+2y^4}$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the denominator.
| Doing the change to polar coordinates:
$$
\frac{x^3y - xy^3}{x^4 + 2y^4} =
\frac
{r^3\cos^3\theta\,r\sin\theta - r\cos\theta\,r^3\sin^3\theta}
{r^4\cos^4\theta + 2r^4\sin^4\theta}
= \frac
{\cos^3\theta\sin\theta - \cos\theta\sin^3\theta}
{\cos^4\theta + 2\sin^4\theta},
$$
dependent of $\theta$ (and independent of $r$), so the limit does not exists.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\cot{142\frac{1}{2}^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
Show that $\cot{142\frac{1}{2} ^\circ} = \sqrt2 + \sqrt3 - 2 - \sqrt6$.
What I have tried:
Let $\theta = 142\frac{1}{2}^\circ \text{ and } 2\theta = 285^\circ$.
$$\cos 285^\circ = \cos 75^\circ$$
$$\cos 75^\circ = \frac{\sqrt3 - 1}{2\sqrt2}$$
$$\cot \theta = \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}}$$
From here, pls help me proceed further. Thank you :)
| Good start. You got $$\cot \theta= \sqrt{\frac{1+\frac{\sqrt3-1}{2\sqrt2}}{1-\frac{\sqrt3-1}{2\sqrt2}}}=\sqrt{\frac{2\sqrt2+\sqrt3-1}{2\sqrt2-(\sqrt3-1)}}\; .$$
To simplify, multiply numerator and denominator by $2\sqrt2+\sqrt3-1.$
Note that the denominator then becomes $8-(\sqrt3-1)^2=(\sqrt3+1)^2. $
Thus $$\cot \theta= -\frac{2\sqrt2+\sqrt3-1}{\sqrt3+1}.$$
Finally, note that $(\sqrt2+\sqrt3)(\sqrt2-1)(\sqrt3+1)=2\sqrt2+\sqrt3-1,$ so
$\cot \theta=-(\sqrt2+\sqrt3)(\sqrt2-1),$ as desired.
| {
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Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$? Question:
Does $\triangle ABC$ exist such that $\triangle ABC \sim \triangle DEF$, with $D, E, F$ being the incentre, centroid, orthocentre of $\triangle ABC$, resp.?
For such a triangle to exist, it must be obtuse. Besides that, I have no idea how to prove or disprove it. For the case of $D, E, F$ being the orthocentre, centroid, circumcentre, it's impossible as they lie on the same line (Euler's line). That's the motivation of the problem. I have a feeling that brute force methods are needed (coordinate geometry). But I hate such an ugly approach. Any idea?
| Here's what I mean by a brute force approach：
Without loss of generality, let's choose an appropriate scale such that $AB=1$.
Set
$$A=(0,0)$$ $$B=(1,0)$$ $$C=(x,y)$$
Then the coordinate of the centroid $E$ is
$$E=\left(\frac {x+1}{3},\frac {y}{3}\right)$$
To find out the coordinate of the incentre $D$, let's calculate the side lengths and perimeter $P$ first
$$|AB|=1$$ $$|AC|=\sqrt{x^2+y^2}$$ $$|BC|=\sqrt{(x-1)^2+y^2}$$ $$P=AB+AC+BC=1+\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}$$
Then the coordinate of the incentre D is
\begin{align} D &=\frac{|BC| A + |AC| B + |AB| C} { P } \\ &= \left (\frac { \sqrt { x ^ { 2 } + y ^ { 2 } } + x} {1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} , \frac{y}{1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} \right) \end{align}
The coordinate of the orthocentre $F(a,b)$ satisfies
$$\left\{ \begin{aligned} \frac{b}{a-1} &=-\frac{x}{y} \\ a&=x \end{aligned} \right.$$
Hence $$F=\left(x, \frac{x-x^2}{y}\right)$$
By similarity
$\frac{DE}{BC}= \sqrt { \frac { \left( \frac { x + 1 } { 3 } - \frac { \sqrt { x ^ { 2 } + y ^ { 2 } + x } } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } } \right) ^ { 2 } + \left( \frac { y } { 3 } - \frac{y}{1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } }} \right) ^ { 2 } } { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }$
$=\frac{EF}{AC}=\sqrt{\frac{\left(\frac { x + 1 } { 3 }-x \right)^2 + \left(\frac { y } { 3 }-\frac { x - x ^ { 2 } } { y } \right)^2}{ x ^ { 2 } + y ^ { 2 } }}$
$=\frac{DF}{AB}=\sqrt{\left(\frac { \sqrt { x ^ { 2 } + y ^ { 2 } } + x } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }-x\right)^2+\left(\frac { y } { 1 + \sqrt { x ^ { 2 } + y ^ { 2 } } + \sqrt { ( x - 1 ) ^ { 2 } + y ^ { 2 } } }-\frac { x - x ^ { 2 } } { y }\right)^2}$
And I've no idea how to solve it with bare hands. Feel free to complete my answer. I'm curious whether there is a purely geometric approach (maybe non-constructive).
| {
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For what values of $a$ and $b$ is the function $\frac{x^ay^b}{x^2+y^2}$ continuous at $(0,0)$? I have the function $$f(x,y)=\begin{cases}\dfrac{x^ay^b}{x^2+y^2} &(x,y)\neq(0,0)\\ 0 &(x,y)=(0,0) \end{cases}$$ I am trying to figure out what constants $a$ and $b$ will make the function continuous at $(0,0)$. I know that the limit has to be $0$ as $(x,y)\to(0,0)$ for it to be continuous.
Using polar coordinates, I think that $a+b \geq 3$ since $x^2+y^2= r^2$ and the limit of any polar function $r^x$ with $x > 0$ as $r\to0$ is $0$. Am I on the right track or am I missing something?
| $0\leq (x-y)^2=x^2+y^2-2xy$
so
$xy\leq \frac{1}{2}(x^2+y^2)$
then
$|\frac{x^ay^b}{x^2+y^2}|\leq|\frac{x^{a-1}y^{b-1}\frac{1}{2}(x^2+y^2)}{x^2+y^2}|=\frac{1}{2}|x^{a-1}y^{b-1}|\to 0$
if $a>1 $ and $b>1$
with the polar coordinate you get a better condition on $a,b$ :
$a+b>2$
| {
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Can anybody help with the integral: $\int \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}} dx$ please? I have searched online. I have got a solution from Wolfram but I don't understand it, or how to reach it. If anyone has a method that would be fantastic, thanks!
| This is an elliptic integral, but it is interesting to point out what is the explicit value of the integral over $(0,1)$.
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \frac{1}{2}\int_{0}^{1}\sqrt{\frac{1-x}{x(1+x)}}\,dx=\frac{1}{2}\int_{0}^{1}\sqrt{\frac{x}{(1-x)(2-x)}}\,dx$$
equals
$$ \int_{0}^{1}\frac{x^2}{\sqrt{(1-x^2)(2-x^2)}}\,dx =\int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{2-\sin^2\theta}}\,d\theta$$
or
$$2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{2-\sin^2\theta}}-\int_{0}^{\pi/2}\sqrt{2-\sin^2\theta}\,d\theta=\sqrt{2}\left[K\left(\tfrac{1}{2}\right)-E\left(\tfrac{1}{2}\right)\right]$$
by representing the complete elliptic integrals according to Mathematica's notation, such that the argument is the elliptic modulus. A series representation for the RHS is given by
$$ \frac{\pi}{\sqrt{2}}\sum_{n\geq 1}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{2n}{1-2n}\cdot\frac{1}{2^n}$$
which due to $\frac{\binom{2n}{n}}{4^n}\approx\frac{1}{\sqrt{\pi n}}$ for large values of $n$, can be approximated by
$$ \frac{\pi}{4\sqrt{2}}+\sqrt{2}\sum_{n\geq 2}\frac{n}{2^n(2n-1)}=\frac{\pi-4}{4\sqrt{2}}+\log(1+\sqrt{2}).$$
The Cauchy-Schwarz inequality provides a simpler (and tighter) upper bound, namely
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx<\sqrt{\int_{0}^{1}(1-x^2)\,dx\int_{0}^{1}\frac{dx}{1+x^2}}=\sqrt{\frac{\pi}{6}}.$$
On the other hand both $K\left(\tfrac{1}{2}\right)$ and $E\left(\tfrac{1}{2}\right)$ can be computed through the $\text{AGM}$ and are explicitly related to the $\Gamma$ function. Indeed the former is given by a multiple of $B\left(\tfrac{1}{4},\tfrac{1}{4}\right)$ and the latter is related to the former via Legendre's relation. This gives
$$ \int_{0}^{1}\sqrt{\frac{1-x^2}{1+x^2}}\,dx = \color{blue}{\frac{\Gamma\left(\tfrac{1}{4}\right)^2}{4\sqrt{2\pi}}-\frac{\pi\sqrt{2\pi}}{\Gamma\left(\tfrac{1}{4}\right)^2}}=\color{red}{\frac{\pi}{2\,\text{AGM}(1,\sqrt{2})}-\frac{\text{AGM}(1,\sqrt{2})}{2}}$$
which can be proved through Fourier-Legendre expansions, too.
| {
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"timestamp": "2023-03-29T00:00:00",
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easy riemann sum problem goes hard Ok, so here it is the integral
$$\int_{0}^{\pi/2}\left[1 + \sin\left(x\right)\right]\,\mathrm{d}x
$$
which I must compute with the definition. And here is my atempt
$$
\int_{0}^{\pi/2}\left[1 + \sin\left(x\right)\right]\,\mathrm{d}x =
\lim_{n \to \infty}\sum_{k = 1}^{n}\mathrm{f}\left(\xi_{k}\right)
\Delta x =
\lim_{n \to \infty}\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n}$$
and this limit is something strange cause I can't solve it. Is this correct up until here ?. thx.
| You are looking to evaluate the limit $$\lim_{n \to \infty}\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n}.$$ Time to simplify! [Warning: this is longer than I expected.] The sum can be split into two parts so that
$$\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n} = \sum_{k = 1}^{n}\frac{\pi}{2n} + \frac{\pi}{2n} \sum_{k = 1}^{n}\sin\left(\frac{k\pi}{2n}\right).$$
The first sum on the right is just $\dfrac \pi 2$. The second sum requires a little bit of trigonometry. The trick is to multiply by a "catalyst". For any number $0 < x < \pi$ you have
$$\sum_{k=1}^n \sin kx = \frac 1 {\sin \frac x2} \sum_{k=1}^n \sin kx \sin \frac x2.$$
The product-to-sum formula from trig states that
$$\sin A \sin B = \frac 12 \left[ \cos(A-B) - \cos(A+B) \right].$$ This means
$$\sin kx \sin \frac x2 = \frac 12 \left[ \cos(k-\tfrac 12)x - \cos(k+\tfrac 12)x \right]$$ so when you add you can take advantage of telescoping:
$$\sum_{k=1}^n \sin kx \sin \frac x2 = \frac 12 \sum_{k=1}^n \left[ \cos(k-\tfrac 12)x - \cos(k+\tfrac 12)x \right] = \frac 12 \left[ \cos \tfrac x2 - \cos(n+ \tfrac 12)x \right].$$ This gives you at last
$$\sum_{k=1}^n \sin kx = \frac 12 \frac{\cos \tfrac x2 - \cos(n+ \tfrac 12)x}{\sin \tfrac x2}.$$ Evaluate this with $x = \dfrac{\pi}{2n}$ to find
$$ \frac{\pi}{2n} \sum_{k=1}^n \sin \left(\frac{k\pi}{2n} \right) = \frac{ \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \left[ \cos \tfrac{\pi}{4n} - \cos \tfrac{(2n+1)\pi}{4n} \right]$$
and finally
$$\sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n} = \frac \pi 2 + \frac{ \frac{\pi}{4n}}{\sin \frac{\pi}{4n}} \left[ \cos \tfrac{\pi}{4n} - \cos \tfrac{(2n+1)\pi}{4n} \right].$$
As you know, $y = \cos x$ is a continuous function and $\frac{\sin x}{x} \to 1$ as $x \to 0$. Thus the limit evaluates to
$$ \lim_{n \to \infty} \sum_{k = 1}^{n}
\left[1 + \sin\left(\frac{k\pi}{2n}\right)\right]\frac{\pi}{2n} = \frac \pi 2 + 1 \cdot [\cos 0- \cos \tfrac \pi 2] = \frac \pi 2 + 1.$$
| {
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Factor $X^4 + 3$ into irreducible factors in $F_7[X]$ I am not quite sure where to start with this problem. I am new to polynomial rings and want to learn how to factor polynomials in polynomial rings made of fields.
Factor $X^4 + 3$ into irreducible factors in $F_7[X]$.
| You are working in $\Bbb F_7[x]$, the ring of polynomials with coefficients in the finite field $\Bbb F_7$. As Will Jagy also suggested, $3\equiv -4$ mod $7$, hence your polynomial is basically $x^4-4$.
$$x^4-4=(x^2+2)(x^2-2)$$
We have other two factors to check. Notice that $3^2=9\equiv 2$ mod $7$, hence the second factor splits in $(x+3)(x-3)$, that is
$$x^4+3=x^4-4=(x^2-2)(x^2+2)=(x+3)(x-3)(x^2+2).$$
Finally, consider $(x^2+2)$. You can check by hand that no element in $\Bbb F_7$ is a solution of $q(x)=x^2+2$; namely, for any $a\in \Bbb F_7,$ we have that $a^2+2\not\equiv 0$ mod $7$. Thus it is irreducible.
For these reasons, the total factorization of $x^4+3$ in $\Bbb F_7[x]$ is
$$x^4+3=(x+3)(x-3)(x^2+2)$$
| {
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Prove $|\alpha-i|=|\beta-i|$ if $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2e^{i\theta}$
If $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2(\cos\theta+i\sin\theta),$ $0<\theta<\pi$, then prove that $|\alpha-i|=|\beta-i|$
My Attempt
$$
z^2-2e^{i\theta}z+1=0\implies\alpha+\beta=2e^{i\theta}\quad\&\quad\alpha\beta=1\\
\alpha,\beta=\frac{2e^{i\theta}\pm\sqrt{4e^{2i\theta}-4}}{2}=e^{i\theta}\pm\sqrt{e^{2i\theta}-1}=e^{i\theta}\pm\sqrt{\cos2\theta-1+i\sin2\theta}\\
=e^{i\theta}\pm\sqrt{-2\sin^2\theta+2i\sin\theta\cos\theta}=e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{-\sin\theta+i\cos\theta}\\
=e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{\cos(\tfrac{\pi}{2}+\theta)+i\sin(\tfrac{\pi}{2}+\theta)}\\
=e^{i\theta}\pm\sqrt{2\sin\theta}.\Big[{\cos(\tfrac{\pi}{4}+\tfrac{\theta}{2})+i\sin(\tfrac{\pi}{4}+\tfrac{\theta}{2})}\Big]\\
$$
As it was asked as a multiple choice question with the solution being one of the options, I think my attempt seems to be more complicated. So what is suggested to be the easiest way to find the solution ?. Can I use geometry ?
| This problem can be worked out using polar coordinates, but it seems to be a bit simpler to use rectangular coordinates.
That $|\alpha-i|=|\beta-i|$ follows from proving this claim: If $z$ is any root of $z+\frac{1}{z}=2e^{i\theta}$ with $0 \lt \theta \lt \pi$, then $|z-i| = \sqrt{2}$.
Let $z = x + yi$ and let $2e^{i\theta} = a + bi$, so $4 = a^2 + b^2$ and $b \gt 0$. The quantity $x^2 + y^2$ will appear often, so abbreviate it as $K$.
With these substitutions, $z+\dfrac{1}{z}=2e^{i\theta}$ becomes $x + yi + \dfrac{x - yi}{K} = a + bi$. Thus $a = x(1 + 1/K)$ and $b = y(1 - 1/K)$. Substituting these into $4 = a^2 + b^2$, we find
$$
\begin{align}
4 &= x^2(1 + 2/K + 1/K^2) + y^2(1 - 2/K + 1/K^2)\\
&= (x^2 + y^2) + 2(x^2 - y^2)/K + (x^2 + y^2)/K^2\quad\\
&= K + 2(K - 2y^2)/K + 1/K.\\
\end{align}
$$
After multiplying by $K$ and rearranging, this becomes
$$
4y^2 = K^2 - 2K + 1.
$$
Clearly, $2y = \pm(K - 1)$, but which sign is correct? Since $b = y(1 - 1/K)$, we have $bK = y(K - 1)$. The left side is positive, so $y$ and $K - 1$ must have the same sign.
Thus we have $2y = x^2 + y^2 - 1$, or $2 = x^2 + (y-1)^2$. This defines a circle of radius $\sqrt{2}$ centered at $(0,1)$. In complex terms, the distance from $z$ to $i$ is $\sqrt{2}$, as claimed.
| {
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Error in fun problem of series and the binary representation of n The problem is as follows:
Let $a_n = :$ number of digits of $n$ in binary notation, and $$A_n = \frac{1}{n} \sum_{k=1}^{n} (-1)^{a_n} $$.
Show that $\limsup A_n = 1/3$.
From previous questions, I have asserted that
$$a_n =
\left\{
\begin{array}{111111}
1 & \mbox{if } n = {1}\\
2 & \mbox{if } n = {2,3}\\
3 & \mbox{if } n = {4,...,7} \\
4 & \mbox{if } n = {8,...,15} \\
...\\
k+1 & \mbox{if } n = {2^k,...,(2^{k+1} - 1)}
\end{array}
\right.$$
and that $A_n$ increases for $k+1$ even and decreases for $k+1$ odd. Trying to solve the problem at hand, I wrote:
The terms of $a_n$ equal to a particular $k+1$ are those of order $2^k$ to $2^{k+1}$. Hence there are $2^{k+1} - 2^k = 2^k$ such terms.
We will call upper peaks all terms $A_i$ such that $A_{i-1} \leq A_i \geq A_{i+1}$. This means that $a_i$ is even, while $a_{i+1}$ is odd, which, from the definition of $a_n$ happens if $i = 2^{k+1} - 1$, for some k odd. Then,
$$A_i = \frac{1}{2^{k+1} - 1} \sum_{k=1}^{n} (-1)^a_n$$
Explicitly computing the sum, as there are $2^{k+1} - 2^k = 2^k$ terms equal to a particular $k+1$,
$$A_i = \frac{1}{2^{k+1} - 1} (-1 +1 +1 -... +1)= \frac{1}{2^{k+1} - 1} (-2^0 + 2^1 - 2^2...+2^{k})$$
Notice that we can pair every two terms of the sum into a pair of the form $2^{k+1} - 2^k = 2^k$. Hence,
$$A_i = \frac{1}{2^{k+1} - 1} (2^0 + 2^2 + ... + 2^{k-1})$$
Now the sum is but a simple geometric progression, of ratio $2^2/2^0 = 2^2$, and with $k-1-0+1 = k$ terms. By the formula of the sums of geometric progressions,
$$A_i = \frac{1}{2^{k+1} - 1} \times \frac{1 - {2^{2}}^{k/2}}{1 - 2^2} = \frac{1 - 2^k}{3 - 3*2^{k+1}} $$
The issue here is that this $A_i$ has limit of $1/6$, and not $1/3$, as desired.
Hope you can help me, I found this a pretty fun problem to try and solve.
| In the last line, instead of
$$A_i = \frac{1}{2^{k+1} - 1} \times \frac{1 - {2^{2}}^{k/2}}{1 - 2^2} $$
it should be
$$A_i = \frac{1}{2^{k+1} - 1} \times \frac{1 - 2^{k+1}}{1 - 2^2}=\frac13$$
| {
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Real matrix satisfying $A^3=4I_n-3A$
Let $A\in M_n(\mathbb{R}) $ so that $A^3=4I_n-3A$. Prove that $\det(A+I_n) =2^n$.
My work : $A$ 's eigenvalues are the roots of $x^3+3x-4=0$, so one of the eigenvalues is $1$ and the others are $\lambda_1$ and $\lambda_2$, the roots of $x^2+x+4=0$.
Hence,
$$\det(A+I_n) =2(\lambda_1+1)(\lambda_2+1)=2(\lambda_1 \lambda_2 +\lambda_1+\lambda_2+1) =2(4-1+1)=8$$ What is my mistake?
| Careful, you only know that the spectrum of $A$ is contained in the set of zeroes of $x^3+3x-4$.
Since $A$ is a real matrix, its minimal polynomial has real coefficients and divides $x^3+3x-4$, so it is one of
$$x-1, x^2+x+4, (x-1)(x^2+x+4)$$
Hence the characteristic polynomial is of the form $\chi_A(x) = (x-1)^{n-2r}(x^2+x+4)^{r}$ for some $0 \le r \le \frac{n}2$.
Therefore if we denote $\lambda_1, \lambda_2$ the zeroes of $x^2+x+4$, we have
$$\det(A+I_n) = (1+1)^{n-2r}(\lambda_1+1)^r(\lambda_2+1)^r = 2^{n-2r}(\lambda_1\lambda_2+\lambda_1+\lambda_2+1)^r = 2^{n-2r}4^r = 2^n$$
| {
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Proving $3\mid p^3 \implies 3\mid p$ I want to prove $3\mid p^3 \implies 3\mid p$ (Does it?)
The contrapositive would be $3 \nmid p \implies 3 \nmid p^3$ I believe.
$3\nmid p \implies p = 3q + r$ ($0<r<3$), so $p^3 = 27q^3+27q^2r+9qr^2+r^3$
Dividing by $3$ we get $3(9q^3 + 9q^2r + 3qr^2) + r^3$
Is this correct so far? How do I finish the proof please? Do I need to show that $r^3$ can never be a multiple of $3$?
| Since $0\lt r\lt3$ so either $r=1$ or $r=2$.
If $r=1$ then $r^3=1$ . So $3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+1$ which leaves remainder $1$ when divided by $3$
And similarly, if $r=2$ then $r^3=8$ . So
$3(9q^3+9q^2r+3qr^2)+r^3=3(9q^3+9q^2r+3qr^2)+8=3(9q^3+9q^2r+3qr^2)+3\cdot2+2=3(9q^3+9q^2r+3qr^2+2)+2$
which leaves remainder $2$ when divided by $3$.
It follows that both of the two expressions are not divisible by $3$ and you reach to a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the following sequence converges. Please Critique my proof. The problem is as follows:
Let $\{a_n\}$ be a sequence of nonnegative numbers such that
$$
a_{n+1}\leq a_n+\frac{(-1)^n}{n}.
$$
Show that $a_n$ converges.
My (wrong) proof:
Notice that
$$
|a_{n+1}-a_n|\leq \left|\frac{(-1)^n}{n}\right|\leq\frac{1}{n}
$$
and since it is known that $\frac{1}{n}\rightarrow 0$ as $n\rightarrow \infty$. We see that we can arbitarily bound, $|a_{n+1}-a_n|$. Thus, $a_n$ converges.
My question:
This is a question from a comprehensive exam I found and am using to review.
Should I argue that we should select $N$ so that $n>N$ implies $\left|\frac{1}{n}\right|<\epsilon$ as well?
Notes: Currently working on the proof.
| Define $b_k := a_{2k+1}$. Then
$$b_k \leq a_{2k} + (-1)^{2k}\frac{1}{2k} \leq b_{k-1} + (\frac{1}{2k} - \frac{1}{2k-1}) \leq b_{k-1}$$
Since $b_k$ is non-negative and non-increasing: $b_k \to b$.
Suppose $a_n \nrightarrow b$. Then there exists an $\varepsilon > 0 $ s.t. for infinitely many $n$ holds $|a_{2n} - b| > \varepsilon$.
Assume that $|a_{2m+1}-a_m| > \frac{\varepsilon}{2}$ for infinitely many $m$. Then, since $a_{2m+1}- a_m \leq \frac{1}{2m}$ we have that
\begin{align}
a_{2m+1} - a_m < - \frac{\varepsilon}{2}
\end{align}
for infinitely many $m$. Let $M := \{m \geq 1 : a_{2m+1} - a_m < - \frac{\varepsilon}{2} \text{ is fulfilled for } m \}$
\begin{align*}
d_m := 1_M (m)
\end{align*}
This implies
\begin{align*}
0 \leq a_{2m+1} = a_1 + \sum_{k=1}^{2m} (a_{k+1} - a_k ) = a_1 + \sum_{k=1}^m (a_{2k+1} - a_{2k}) + \sum_{k=1}^m (a_{2k} - {a_{2k-1}}) \\
\leq a_1 + \sum_{k=1}^m (-1)^{2k} \frac{1}{2k}- \frac{\varepsilon}{2} d_k + \sum_{k=1}^m (-1)^{2k-1}\frac{1}{2k-1} \to a_1 - \sum_{k=1}^\infty \frac{\varepsilon}{2} d_k + \sum_{i=1}^\infty (-1)^i \frac{1}{i} = - \infty
\end{align*}
since $|M| = \infty$ and the last series converges. This is a contradiction.
Therefore we have that there exists $K\geq 1$ s.t. for all $k\geq K$ it holds: $|a_{2k+1} - a_k| \leq \frac{\varepsilon}{2}$. We can conclude that
\begin{align*}
|a{2n+1} - b| \geq |a_{2n} - b| - |a_{2n+1} - a_n| \geq \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2}
\end{align*}
for infinitely $n \geq K$. Contradiction. Thus $a_n \to b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove that in every triangle the inequality $a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $ takes place Prove that in every triangle the inequality $$a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $$ takes place, with the usual notations ($a,b,c$ lengths of sides, $r_a, r_b, r_c$ radii of corresponding excircles, $R$ radius of circumscribed circle and $r$ radius of inscribed circle)
I've tried to solve it using Chebyshev (Rearrangement Inequality) or Jensen Inequality (for $f(x)=\frac{x^3}{p-x}$), however I always end up minoring the inequality too much, or obtaining an inequality which I am not able to prove.
| Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x>0$, $y>0$ and $z>0$ and in the standard notation we need to prove that
$$\sum_{cyc}\frac{a^3\cdot2S}{b+c-a}\geq8S\left(\frac{abc}{2S}-\frac{2S}{a+b+c}\right)^2$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}\geq\frac{(4abc(a+b+c)-16S^2)^2}{16S^2(a+b+c)^2}$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}\geq\frac{\left(4abc-\prod\limits_{cyc}a+b-c)\right))^2}{(a+b+c)\prod\limits_{cyc}(a+b-c)}$$ or
$$(x+y+z)\sum_{cyc}(x+y)^3xy\geq2\left(\sum_{cyc}(x^2y+x^2z)\right)^2$$ or
$$\sum_{cyc}(x+y)\sum_{cyc}(x+y)^3xy\geq4\left(\sum_{cyc}(x^2y+x^2z)\right)^2.$$
Now, by C-S
$$\sum_{cyc}(x+y)\sum_{cyc}(x+y)^3xy\geq\left(\sum_{cyc}(x+y)^2\sqrt{xy}\right)^2.$$
Id est, it's enough to prove that
$$\sum_{cyc}(x+y)^2\sqrt{xy}\geq2\sum_{cyc}(x^2y+x^2z)$$ or
$$\sum_{cyc}(x+y)^2\sqrt{xy}\geq2\sum_{cyc}(x^2y+xy^2)$$ or
$$\sum_{cyc}(x+y)\sqrt{xy}(\sqrt{x}-\sqrt{y})^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Harmonic series in probability mass function problem Suppose $X$ is a discrete random variable with possible values $\{1, 2, 3,\dots\}.$ Further, suppose the p.m.f is $$c\left(\frac{1}{x}-\frac{1}{x+1}\right)\enspace\text{s.t. $c > 0$}$$
Find c and $E[X].$
Idea:
We have $$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\left(\sum_{x=1}^{\infty}\frac{1}{x}-\sum_{x=1}^{\infty}\frac{1}{x+1}\right)$$
But since $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series, diverges. Thus, there is no value for $c$.
Since it diverges, $E[X]$ does not exist.
Questions:
Is it possible for c not to exist? Did I do a mistake?
Update:
$$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\sum_{x=1}^{\infty}\left(\frac{1}{x}-\frac{1}{x+1}\right)$$
by telescoping series we have
$$1=c\cdot 1$$
So, our p.m.f is $$\left(\frac{1}{x}-\frac{1}{x+1}\right)$$
But, $$E[X]=1\cdot\frac{1}{2}+2\cdot\frac{1}{6}+3\cdot\frac{1}{12}+4\cdot \frac{1}{20}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots$$
But that diverges. So, $E[X]$ doesn't exist.
| You can write $$\sum a_n-b_n=\sum a_n-\sum b_n$$only if at least one of $\sum a_n$ or $\sum b_n$ is bounded. In this case$$\sum_{x=1}^{\infty}{1\over x}-{1\over x+1}{=\left(1-{1\over 2}\right)\\+\left({1\over 2}-{1\over 3}\right)\\+\left({1\over 3}-{1\over 4}\right)\\+\left({1\over 4}-{1\over 5}\right)\\+\cdots\\=1}$$therefore $$c=1$$ and the rest is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A binomial double sum I was doing some numerical experiment and I found that
$$\sum_{k=2}^{\infty}\frac{1}{k(k-1)}\sum_{n=k}^{\infty}\frac{1}{\binom{n}{k}(n+1)}=\sum_{k=2}^{\infty}\frac{1}{(k-1)k^2}=2-\zeta(2).$$
I wonder if the following variation
$$\sum_{k=2}^{\infty}\frac{1}{k(k-1)}\sum_{n=k}^{\infty}\frac{1}{\binom{n}{k}(n-1)}$$
has a closed form. Is it related to $\zeta(2)$ or $\zeta(3)$?
| We have that the inner sum equals
$$
\eqalign{
& \sum\limits_{k\, \le \,n} {{1 \over {\binom{n}{k}\left( {n - 1} \right)}}} = \cr
& = \sum\limits_{0\, \le \,n} {{1 \over {\binom{n+k}{k}\left( {n + k - 1} \right)}}} = \cr
& = k!\sum\limits_{0\, \le \,n} {{1 \over {\left( {n + k} \right)^{\,\underline {\,k\,} } \left( {n + k - 1} \right)}}} = \cr
& = k!\sum\limits_{0\, \le \,n} {{1 \over {\left( {n + 1} \right)^{\,\overline {\,k\,} } \left( {n + k - 1} \right)}}} \cr}
$$
Therefore
$$
\eqalign{
& S = \sum\limits_{2\, \le \,k} {{1 \over {k\left( {k - 1} \right)}}\sum\limits_{k\, \le \,n} {{1 \over {\binom{n}{k}\left( {n - 1} \right)}}} } = \cr
& = \sum\limits_{2\, \le \,k} {{{k!} \over {k\left( {k - 1} \right)}}\sum\limits_{0\, \le \,n}
{{1 \over {\left( {n + 1} \right)^{\,\overline {\,k\,} } \left( {n + k - 1} \right)}}} } = \cr
& = \sum\limits_{0\, \le \,k} {{{\left( {k + 2} \right)!} \over {\left( {k + 2} \right)\left( {k + 1} \right)}}\sum\limits_{0\, \le \,n}
{{1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 2\,} } \left( {n + 1 + k} \right)}}} } = \cr
& = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k} {{{k!} \over {\left( {n + 1} \right)^{\,\overline {\,k + 2\,} } \left( {n + 1 + k} \right)}}} } = \cr
& = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k} {{{k!\,\left( {n + 1} \right)^{\,\overline {\,k\,} } }
\over {\left( {n + 1} \right)^{\,\overline {\,k + 2\,} } \left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}} } = \cr
& = \sum\limits_{0\, \le \,n} {{1 \over {\left( {n + 1} \right)^{\,2} \left( {n + 2} \right)}}\sum\limits_{0\, \le \,k}
{{{1^{\,\overline {\,k\,} } \,\left( {n + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {n + 3} \right)^{\,\overline {\,k\,} } \left( {n + 2} \right)^{\,\overline {\,k\,} } }}} } = \cr
& = \sum\limits_{0\, \le \,n} {{1 \over {\left( {n + 1} \right)^{\,2} \left( {n + 2} \right)}}
{}_3F_{\,2}\left( {\left. {\matrix{ {1,\;1,n + 1} \cr {n + 3,n + 2} \cr } \;} \right|\;1} \right)} \cr}
$$
Unfortunately, the Hypergeometric does not look to be
of a type expressible in a simpler way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfrac{\cos 9x}{\sin 27x}=0$
*$\dfrac{1}{\cos x\cos 2x}+\dfrac{1}{\cos 2x\cos 3x}+\dfrac{1}{\cos 3x\cos 4x}=0$
These equations have patterns, and I know if we can use the pattern we will solve the equations very easily. I managed to use the pattern on the first equation to find a telescoping series and get this (it is not a full solution but it is the way to solve the first equation):
We have $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ therefore
\begin{align*}
(1)&\Leftrightarrow\dfrac{1}{(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)}=\dfrac18\\
&\Leftrightarrow\left(2\tan x\cdot\dfrac{1}{1-\tan^2x}\right)\cdot\dfrac{1}{1-\tan^22x}\cdot\dfrac{1}{1-\tan^24x}=\dfrac14\tan x\\
&\Leftrightarrow\left(2\tan 2x\cdot\dfrac{1}{1-\tan^22x}\right)\cdot\dfrac{1}{1-\tan^24x}=\dfrac12\tan x\\
\end{align*}
and so on.
However, I can't manage to solve the last four. I can't find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ in the first equation. So here are my questions:
*
*How to solve equations 2, 3, 4, and 5?
*What is the strategy to find the key equalities like $\frac{2\tan x}{1-\tan^2x}=\tan 2x$ to get a telescoping series for each equation?
Thank you in advance.
| $2. 2\cos2y+1=2(1-2\sin^2y)+1=\dfrac{\sin3y}{\sin y}$ for $\sin y\ne0$
$3.\dfrac{\cos2y}{\sin3y}=\dfrac{2\cos2y\sin y}{2\sin3y\sin y}=?$
$4. \dfrac{\cos x}{\sin3x}=\dfrac{\sin(3x-x)}{2\sin3x\sin x}=?$
$5.\sin x=\sin((n+1)x-nx)=?$
Set $n=1,2,3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $(x^2+y^2)^2$ if $x,y$ are real number such that $x^2+2xy-y^2=6$ Then find minimum value of $(x^2+y^2)^2$
what i try : $x^2+2xy+y^2-2y^2=6$ or $(x+y)^2-\bigg(\sqrt{2} y\bigg)^2=6$
put $\displaystyle (x+y)=\sqrt{6}\cos \alpha$ and $\displaystyle \sqrt{2}y=\sqrt{6}\sin \alpha$
$\displaystyle x=\sqrt{6}\cos \alpha-\sqrt{3}\sin \alpha$ and $\displaystyle y =\sqrt{3}\sin \alpha$
$\displaystyle x^2+y^2=3\bigg[\bigg(\sqrt{2}\cos \alpha-\sin \alpha\bigg)^2+\sin^2\alpha\bigg)\bigg]$
$\displaystyle x^2+y^2=3\bigg[2\cos^2\alpha+\sin^2\alpha-2\sqrt{2}\cos \alpha\sin \alpha+\sin^2\alpha\bigg]$
$\displaystyle x^2+y^2=3\bigg(2-\sqrt{2}\sin 2\alpha\bigg)\geq 3(2-\sqrt{2})$
but answer is $\sqrt{18}$ How do i solve it Help me please
| I think a fun way to do the problem similar to what you've done is by using polar co-ordinates. Sub $x=r \cos \theta$ and $y=r \sin \theta$. Then you want to minimise $(x^2+y^2)^2=r^4$. Note that the constraint simplifies massively.
$$ x^2+2xy-y^2=6$$
$$r^2( \cos ^2 \theta +2 \sin \theta \cos \theta -\sin ^2 \theta )=6$$
$$r^2(\cos 2\theta +\sin 2\theta )=6$$
$$r^2 \sqrt{2} \sin ( 2\theta +\frac{\pi}{4})=6$$
$$r^2=3 \sqrt{2} \csc( 2\theta +\frac{\pi}{4})$$
Hence the minimum of $r^4$ is $(3\sqrt {2})^2=18$
| {
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"timestamp": "2023-03-29T00:00:00",
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I'm having a hard time rationalizing the proof that $2^n > n^2$ if $n >4$ I'm not sure how to apply $n > 4$ in the proof, when I work it out I get that $n >2$...
Basis Step $P(5) = 2^5 > 5^2 = 32 > 25 $ which is True
Hypothesis assume $P(k) = 2^k > k^2$ is true. Prove that $P(k+1)$ is true.
Substituting $k+1$ into the hypothesis yields:
$2^{k+1} > (k+1)^2$
$2 \bullet 2^k > k^2 +2k +1$
$2^k > \frac{k^2 +2k +1}{2}$
Since I assume $2^k > k^2$ I then try to find the conditions where $k^2 > \frac{k^2 +2k +1}{2}$
$2k^2 > k^2 +2k +1$
$k^2 -2k - 1 > 0$ which is true when $k > 4$ (and also $k >2$). Therefore the proof is true via mathematical induction?
| First of all, this is false when $n=3$. $2^3 = 8 < 9 = 3^2$. This becomes clear if you properly write out your inequality:
$$2^k > k^2 >\frac{k^2 + 2k + 1}{2}.$$
The right inequality holds for $k > 2$ but the left one only holds for $k =5$, and $k \geq 5$ once you prove the inequality. When you plug in $n=3$, you get
$$2^3 = 8 \ngeq 9 = 3^2 > \frac{4^2}{2} = 8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i solve it Help me please
| In fact, you are given that $$ \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{px^3+qx^8}{(x^5-1)^2}+c$$ Because of the denominator in the integrand, rewrite the rhs as
$$\frac{P_n(x)}{x^5-1}$$ Differentiate both sides to get
$$\frac{2x^7+3x^2}{(x^5-1)^2}=\frac{\left(x^5-1\right) P_n'(x)-5 x^4 P_n(x)}{\left(x^5-1\right)^2}$$ that is to say
$${2x^7+3x^2}={\left(x^5-1\right) P_n'(x)-5 x^4 P_n(x)} $$In order to have $x^7$ implies that $5+(n-1)=7$ that is to say $n=3$. So, let
$$P_3(x)=a+bx+c x^2+dx^3$$ Replace, expand and group terms in the rhs to get
$${2x^7+3x^2}=-b-2 c x-3 d x^2-5 a x^4-4 b x^5-3 c x^6-2 d x^7$$ which implies $b=0$, $c=0$, $d=-1$ and $a=0$. So $P_3(x)=-x^3$. For here
$$px^3+qx^8=-x^3(x^5-1)=x^3-x^8\implies p=1 \qquad \text{and} \qquad q=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify $\sqrt{2+\sqrt{3}}$ $?$
Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$
The answer to this question is $\frac{4}{3}$ in a workbook.
How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2}$ $=$ $\sqrt{(\sqrt2)^2 + 2(\sqrt2)(1) + (1)^2}$ $=$
$\sqrt{(\sqrt2 + 1)^2}$ $=$
$\sqrt2 + 1$
But I can't simplify $\sqrt{2+\sqrt3}$ like that as $2+\sqrt3$ is can't be written as squares of two numbers. Is there any other method?
| Hints:
$$\sqrt 2+\sqrt 6=\sqrt 2(1+\sqrt 3)=\sqrt{2(1+\sqrt 3)^2}=\sqrt{2(4+2\sqrt 3)}=2\sqrt{2+\sqrt 3}$$
$$\sqrt{2+\sqrt 3}=\frac 1{\sqrt 2}\sqrt{4+2\sqrt 3}=\frac 1{\sqrt 2}\sqrt{(\sqrt 3+1)^2}=\frac{\sqrt 3+1}{\sqrt 2}=\frac{\sqrt 6+\sqrt 2}2$$
| {
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"url": "https://math.stackexchange.com/questions/3151235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Prove:$$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
I tried to use the Laplace expansion, but it seems useless.
| Hint:
$C_1'=C_1+C_2+C_3$
$$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}$$
$$=\begin{vmatrix}a+b+c&b&c\\b+c+a&c&a\\c+a+b&a&b\end{vmatrix}$$
$$=(a+b+c)\begin{vmatrix}1&b&c\\1&c&a\\1&a&b\end{vmatrix}$$
Now either expand or
use $R_2'=R_2-R_1, R_3'=R_3-R_1$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$ Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$
Answer: $K=(ab+bc+ca)$
My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3-a^3\end{vmatrix}\\&=\begin{vmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3-a^3\end{vmatrix}\\&=(b-a)(c-a)\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}\\&=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}\\\end{align}$$
I don't know what should I do next, maybe I've made a mistake but I didn't notice it
| Note that we can rewrite the determinant as a sum:
$$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}b^2&c^2\\b^3&c^3\end{vmatrix} - \begin{vmatrix}a^2&c^2\\a^3&c^3\end{vmatrix} + \begin{vmatrix}a^2&b^2\\a^3&b^3\end{vmatrix} = (b^2c^3 - b^3c^2) - (a^2c^3 - a^3c^2) + (a^2b^3 - a^3b^2)$$
Now, we can solve for $K$ by simply solving $K = \frac{(b^2c^3 - b^3c^2) - (a^2c^3 - a^3c^2) + (a^2b^3 - a^3b^2)}{(a - b)(b - c)(c - a)}$. If you need another hint just comment and I will add more.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of the absence of isolated vertices. We have a graph with n vertices. Each edge can appear with probability $\frac{1}{n\ln(n)}$. Prove that $\lim\limits_{n\rightarrow\infty}Pr[$there is no isolated vertices$] = 0$.
I think that first of all we should connect vecrtices so there won't be isolated vertices. We connect them by pairs $\binom{n}{2}\frac{1}{n\ln(n)} + \binom{n-2}{2}\frac{1}{n\ln(n)} + \cdots + \binom{2}{2}\frac{1}{n\ln(n)}$ and divide it on $\frac{n}{2}$ (because connection does not depends on order). And do smth with other edges $\sum_{i = 1}^{\frac{n(n-1)}{2} - \frac{n}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - \frac{n}{2}}$.
So the probability is $\frac{(\sum_{i = 1}^{\frac{n(n-1)}{2} - \frac{n}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - \frac{n}{2} - i})\cdot\frac{(\binom{n}{2}\frac{1}{n\ln(n)} + \binom{n-2}{2}\frac{1}{n\ln(n)} + \cdots + \binom{2}{2}\frac{1}{n\ln(n)})}{\frac{n}{2}}}{\sum_{i = 1}^{\frac{n(n-1)}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - i}}$. I think that it is alright but I am not sure.
| I assume that the edge choices are independent. Put $p=\frac{1}{n\ln(n)}$. A probability that a given vertex is isolated is $$(1-p)^{n-1}\ge 1-(n-1)p\ge 1-\tfrac{1}{\ln n}$$ by Bernoulli's inequality, and the last value tends to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.
| You can prove that $$\color{red}{(x-1)^3\geq {3\over 4}x-1}$$ for all nonegative $x$; it is equivalent to $x(2x-3)^2\ge 0$, so $$(a-1)^3+(b-1)^3+(c-1)^3\geq {3\over 4}(a+b+c)-3= -{3\over 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solve tan(x)+cos(x)=1/2 Is it possible (not numerically) to find the $x$ such as:
$$
tan(x)+cos(x)=1/2
$$
?
All my tries finishes in a 4 degree polynomial. By example, calling c = cos(x):
$$
\frac{\sqrt{1-c^2}}{c}+c=\frac{1}{2}
$$
$$
\sqrt{1-c^2}+c^2=\frac{1}{2}c
$$
$$
1-c^2=c^2(\frac{1}{2}-c)^2=c^2(\frac{1}{4}-c+c^2)
$$
$$
c^4-c^3+\frac{5}{4}c^2-1=0
$$
| If we put $t=x/2$ then we get $${2\tan t\over 1-\tan ^2t} +2\cos ^2 t -1={1\over 2}$$
Let $y= \tan t$. Since $\cos ^2t = {1\over 1+y^2}$ we get $$3y^4+4y^3-4y^2+4y+1=0$$
which I'm not sure if any helps. :(
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the type of the equation Given
$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0$
I need to find the type of the equation, for that i tried to get the canonical form
So, turn into lambdas
$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3 = 0$
from here i can't get the form which is looks like
$\pm\xi^2 \pm \eta^2 \pm \zeta^2 =0$
any help or advice is much appreciated
| The equation
$$\sum_{i,j=1}^na_{ij}u_{x_ix_j}+a(x,u,\nabla u)=0$$
can be reduced to canonical form by applying the non singular linear transformation $\xi=B^Tx$, where B is a matrix such that the transformation $y=B\eta$ reduces the quadratic form $\sum_{i,j=1}^na_{ij}y_iy_j$ to canonical form.
Example. Reduce to canonical form
$$4u_{xx} +2u_{yy} - 6u_{zz} +6u_{xy} + 10u_{xz} +4u_{yz} + 2u =0.$$
$$4\lambda_1^2 + 2\lambda_2^2 - 6\lambda_3^2 + 6\lambda_1\lambda_2 + 10\lambda_1\lambda_3 + 4\lambda_2\lambda_3=
\frac{(4\lambda_1+3\lambda_2+5\lambda_3)^2}{4}-\frac{(\lambda_2+7\lambda_3)^2}{4}$$
Let
$$y_1=\frac{4\lambda_1+3\lambda_2+5\lambda_3}{2},\;y_2=\frac{\lambda_2+7\lambda_3}{2},\;y_3=\lambda_3$$
$\Rightarrow$
$$\lambda_1=\frac12y_1-\frac32y_2+4y_2,\;\lambda_2=2y_2-7y_2,\;\lambda_2=y_3$$
$\Rightarrow$
$$B^T=\begin{pmatrix}\frac{1}{2} & -\frac{3}{2} & 4\\
0 & 2 & -7\\
0 & 0 & 1\end{pmatrix},\quad B=\begin{pmatrix}\frac{1}{2} & 0 & 0\\
-\frac{3}{2} & 2 & 0\\
4 & -7 & 1\end{pmatrix}$$
Transform to canonical form is:
$$\xi=\frac{x}{2},\;\eta=-\frac{3}{2}x+2y,\;\zeta=4x-7 y+z$$
After reducing we get
$$u_{\xi\xi}-u_{\eta\eta}+2u=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the value of $\lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$
Finding the value of $\displaystyle \lim\limits_{n\rightarrow \infty}\sqrt{n}\int^{\frac{\pi}{4}}_{0}\cos^{2n-2}(x)\mathrm dx$
What I tried
Let $\displaystyle I_{k} =\int^{\frac{\pi}{4}}_{0}\cos^{k}(x)\mathrm dx=\int^{\frac{\pi}{4}}_{0}\cos^{k-1}(x)\cdot \cos (x)\mathrm dx$
$$ I_{k}=\left.\cos^{k-1}(x)\sin x\right|^{\frac{\pi}{4}}_{0}+(k-1)\int^{\frac{\pi}{4}}_{0}\cos^{k-2}(x)\left(1-\cos^2 x\right)\mathrm dx$$
$$ k I_{k}=\bigg(\frac{1}{2}\bigg)^{\frac{k}{2}}+(k-1)I_{k-2}$$
How do I solve it? Help me please.
| Result
Let
$$f(n) = \int_0^\frac{\pi}{4} \cos(x)^n\,dx$$
then
$$\lim_{n\to \infty } \, \sqrt{n} f(n)=\sqrt{\frac{\pi }{2}} \simeq 1.2533141373155001\tag{1}$$
and
$$\lim_{n\to \infty } \, \sqrt{n} f(2n)=\frac{\sqrt{\pi }}{2} \simeq 0.8862269254527579\tag{2}$$
Here $(2)$ is the limit asked for in the OP.
Derivation
Decomposing the integration region we write
$$f(n) = i_1(n) - i_2(n)$$
where
$$i_1(n) = \int_0^\frac{\pi}{2} \cos(x)^n\,dx$$
$$i_2(n) = \int_\frac{\pi}{4}^\frac{\pi}{2} \cos(x)^n\,dx$$
Because in the integration interval of $i_2$ we have $\cos(x) \le \cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$ we see immediately that $i_2$ goes to zero exponentially for $n\to\infty$.
Hence we focus on $i_1$.
Changing variables $x\to\arctan(t)$, $dx\to\frac{1}{1+t^2}$ gives
$$i_1(n) = \int_0^{\infty } \left(t^2+1\right)^{-\frac{n}{2}-1} \, dt$$
then $t\to \frac{1}{\sqrt{r}}$ gives
$$i_1(n) = \frac{1}{2} \int_0^{\infty } r^{-\frac{3}{2}} \left(\frac{1}{r}+1\right)^{-\frac{n}{2}-1} \, dr$$
And finally $r\to \frac{1}{z}-1$ leads to
$$i_1(n) = \frac{1}{2} \int_0^1 z^{-\frac{1}{2}} (1-z)^{\frac{n}{2}-\frac{1}{2}} \, dz \\
= \frac{1}{2} B(\frac{1}{2},\frac{n+1}{2})=\frac{1}{2}
\frac{\Gamma(\frac{1}{2}) \Gamma(\frac{1+n}{2})} {\Gamma(1+\frac{n}{2})}\tag{3}$$
The asymptotic behavour is then found using Stirling's formula to give the results announced in the beginning.
Notice that for (2) we have to let $n\to 2n$ in $(3)$ before taking the limit.
Remark
Herewith we have also found that for the more general integral we have for any positive $k$ that
$$\lim_{n\to \infty } \, \sqrt{k n}\int_0^\frac{\pi}{4} \cos(x)^{k n}\,dx = \sqrt{\frac{\pi}{2}}\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Meaning of linear independence with row vectors So far I have understood that a set of vectors $S = {v_1, v_2, . . . , v_k }$ in a vector space V is linearly independent
when the vector equation
$c_1v_1 + c_2v_2 + . . . + c_kv_k = 0$
has only the trivial solution$c_1 = 0, c_2 = 0, . . . , c_k = 0.$
An example in matrix form is:
$\begin{bmatrix}1 & 1 & 2 & 4 \\
0 & -1 & -5 & 2 \\
0 & 0 & -4 & 1 \\
0 & 0 & 0 & 6 \\
\end{bmatrix}
\begin{bmatrix}
c_1\\
c_2 \\
c_3 \\
c_4 \\
\end{bmatrix} =
\begin{bmatrix}
0\\
0 \\
0 \\
0 \\
\end{bmatrix}
$
But a matrix of this form
$\begin{bmatrix}1 & 1 & 2 & 4 \\
0 & -1 & -5 & 2 \\
0 & 0 & -4 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
c_1\\
c_2 \\
c_3 \\
c_4 \\
\end{bmatrix} =
\begin{bmatrix}
0\\
0 \\
0 \\
0 \\
\end{bmatrix}
$
is linearly dependent because it has more than a trivial solution
However, I am confused about row vectors, specifically the idea that to get a basis for a subspace using row vectors we must put the matrix in reduced row echelon form to find the linearly independent vectors. For example here the accepted answer gives an example of finding a basis with row vectors using this
$\begin{bmatrix}1 & 1 & 2 & 4 \\
2 & -1 & -5 & 2 \\
1 & -1 & -4 & 0 \\
2 & 1 & 1 & 6 \\
\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\
0 & -3 & -9 & -6 \\
0 & -2 & -6 & -4 \\
0 & -1 & -3 & -2 \\
\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 1 & 2 & 4 \\
0 & -3 & -9 & -6 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$
and then goes on to say that "Only two of the four original vectors were linearly independent." In what respect are these two vectors linearly independent? This looks exactly like the second example that I gave in which the vectors were dependent because they had more than a trivial solution? Does linear independence with regard to row vectors mean something else? Or does this also only have a trivial solution, and if so, how?
| Yes, it is "exactly like the second example" you gave. In that example, you were showing that the original four vectors were dependent. But there will always be some subset of those vectors that is independent. (That subset might consist of a single vector. A set containing just one vector is always "independent".) In this last problem you are looking for that independent subset.
Your "four vectors" are the rows of the matrix (1, 1, 2, 4), (2, -1, -5, 2), (1, -1, -4, 0), and (2, 1, 1, 6). To determine whether they are "independent" or "dependent", you look for numbers, a, b, c, d, such that a(1, 1, 2, 4)+ b(2, -1, -5, 2)+ c(1, -1, -4, 0)+ d(2, 1, 1, 6)= (a+ 2b+ c+ 2d, a- b- c+ d, 2a- 5b- 4c+ d, 4a+ 2b+ 6d)= (0, 0, 0, 0). So we have the four equations, a+ 2b+ c+ 2d= 0, a- b- c+ d= 0, 2a- 5b- 4c+ d= 0, and 4a+ 2b+ 6d= 0. Adding the first two equations eliminates c: 2a+ b+ 3d= 0. But notice that the fourth equation, 4a+ 2b+ 6d= 0, is just two times that so is not really a new equation. If we add the third equation to 4 times the first equation we also eliminate c: 6a+ 3b+ 9d= 0. We can divide by 3 to get 2a+ b+ 3d= 0.
We have two equations, 3a+ b+ 3d= 0 and 2a+ b+ 3d= 0. Subtracting the second of those from the first gives a= 0. But then we have b+ 3d= 0 so b= -3d but d can be anything. Putting a= 0, b= -3d into a+ 2b+ c+ 2d= 0 we have -6d+ c+ 2d= c- 4d= 0 so c= 4d. That means that a(1, 1, 2, 4)+ b(2, -1, -5, 2)+ c(1, -1, -4, 0)+ d(2, 1, 1, 6)= (0, 0, 0, 0) can be written -3d(2, -1, -5, 2)+ 4d(1, -1, -4, 0)+ d(2, 1, 1, 1, 0)= (0, 0, 0, 0). Then d(2, 1, 1, 1 0)= 3d(2, -1, -5, 2)- 4d(1, -1, -4, 0) or, dividing by d, (2, 1, 1, 0)= 3(2, -1, -5, 2)- 4(1, -1, -4, 0). So those 4 vectors can actually be written in terms of these last two. These two vectors form an independent subset of the original four vectors. The space they span is actually two dimensional, not four dimensional. Using "row reduction" with a matrix is just a more convenient way of doing that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An integration that wolfram cannot help me. $$\int e^{x\sin x+\cos x}\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2x}dx$$
I noted the fact that $\frac{d(x\cos x)}{dx}=-x\sin x+\cos x$ but I cannot apply the substitution on it.
| Let's write an antiderivative Ansatz $f(x)\exp (x\sin x+\cos x)$ so $$f^\prime(x) + f(x)x\cos x=x^2\cos x-\frac{1}{x}\sec x\tan x+\frac{1}{x^2}\sec x\\=x^2\cos x-\left(\frac{1}{x}\sec x\right)^\prime=1+x\cos^2 x-\left(\frac{1}{x}\sec x\right)^\prime-\frac{1}{x}\sec x\cdot x\cos x.$$But by inspection, this has solution $f(x)=x-\frac{1}{x}\sec x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Using transformation to evaluate double integral
Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
$\iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.
Computing the Jacobian of the transformation, I get $\det[DT(x,y)] = 2$. Change the coordinates of the vertices respectively, we get
*
*$A(1,1)$ becomes $(0,2)$
*$B(2,2)$ becomes $(0,4)$
*$C(-1,5)$ becomes $(-6,4)$
*$D(-2,4)$ becomes $(-6,2)$
Solving the double integral, I did:
$$
\int_2^4 \int_{-6}^0 2\left[(x-y)^2-(x+y)^2\right]dxdy
$$
which gave me an answer of 864, however, I should be getting $-54$ as given in the answer sheet, may I know what went wrong?
| HINT
One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
\begin{split}
A &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[\left(\frac{u+v}{2}\right)^2 + \left(\frac{v-u}{2}\right)^2\right] 2\ du\ dv \\
&= \frac12
\int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[(u+v)^2 + (v-u)^2\right] du\ dv \\
&= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[u^2+v^2\right] du\ dv
\end{split}
$$
UPDATE
$$
\begin{split}
A &= \int_{v=2}^{v=4} \int_{u=-6}^{u=0} \left[u^2+v^2\right] du\ dv \\
&= 2 \int_{u=-6}^{u=0} u^2 du
+ 6\int_{v=2}^{v=4} v^2 dv \\
&= \left.\frac23 u^3 \right|_{u=-6}^{u=0}
+ \left.2 v^3 \right|_{v=2}^{v=4} \\
&= -\frac23 (-6)^3 + 2\left(4^3-2^3\right)\\
&= 36\times 4+2(64 - 8)\\
&= 256.
\end{split}
$$
The arithmetic agrees with Wolfram Alpha.
As for your hint, you are (originally) integrating
$$
\iint_A \left(x^2+y^2\right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$. $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdot\cdot\frac{n^3-1}{n^3+1}$
But I could'nt find any pattern in this.So how the limit can be evaluated?
| There's some telescoping going on here. We have
$$\frac{n^3-1}{n^3+1}=\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}.$$
Then
$$\prod_{n=2}^N\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{N-1}{N+2}$$
and
$$\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{N^2+N+1}{N^2-N+1}.$$
Both these products telescope to give a closed form for
$$\prod_{n=2}^N\frac{n^3-1}{n^3+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Geometric proof of $\sin x \geq x - x^3 /6 $? We know (from Taylor expansion for example) that if $x \geq 0$, then $\sin x \geq x - \frac{x^3}{6}$.
In Prove that: $\sin(x) \cos(x) \geq x-x^3$ a geometric proof of the inequality $\sin x \geq x - \frac{x^3}{4}$ is given. Is there any geometric proof of the first one (which is slightly stronger ?
| You meant this holds for acute $x$. Take a radius-$\sqrt{2}$ centre-$O$ circle of radii $OA,\,OB$ with $\angle AOB=x$. We'll work in Cartesian coordinates $X,\,Y$; rotate the diagram so the line segment $AB$, which is of length $2\sqrt{2}\sin\frac{x}{2}$, has endpoints at $$X=\pm\sqrt{2}\sin\frac{x}{2},\,Y=0,$$and let $M$ be the midpoint of the arc $AB$. The circular segment has area $x-\sin x$, which we wish to prove $\le\frac{1}{6}x^3$. We can asymptotically approximate the arc as a parabola passing through the aforementioned endpoints and $$M=\left(0,\,\sqrt{2}\left(1-\cos\frac{x}{2}\right)\right)=\left(0,\,2\sqrt{2}\sin^2\frac{x}{4}\right).$$The parabola has equation $$Y=\frac{\sqrt{2}\sin^2\frac{x}{4}}{\sin^2\frac{x}{2}}\left(2\sin^2\frac{x}{2}-X^2\right),$$so the area under the parabola is $$\frac{2\sqrt{2}\sin^2\frac{x}{4}}{\sin^2\frac{x}{2}}\frac{4\sqrt{2}\sin^3\frac{x}{2}}{3}=\frac{16}{3}\sin^2\frac{x}{4}\sin\frac{x}{2}.$$(Admittedly I had to use $\int_0^a (a^2-X^2)dX=\frac{2a^3}{3}$ there, but geometrically this is equivalent to a pyramid's volume, so look up your favourite "classical" proof of that.) Asymptotically this is approximately, but less than, $$\frac{16x^3}{3\times 4^2\times 2}=\frac{x^3}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fractions, disagreement with Wolfram Alpha On my math homework, I have this problem, and WolframAlpha says that $A=-\frac{1}{7}$ and $B=\frac{1}{7}$. However, while solving the problem on my own, I found a restriction that $A \neq -B$. How is it that this answer works? The problem in question is:
$$\dfrac{1}{x^2 - 3 x - 10} = \dfrac{A}{x+2} + \dfrac{B}{x - 5}$$
| To find $A$ and $B$:
Note that
$(x + 2)(x - 5) = x^2 -3x - 10; \tag 1$
then from
$\dfrac{1}{x^2 -3x - 10} = \dfrac{A}{x + 2} + \dfrac{B}{x - 5} \tag 2$
we have
$A(x - 5) + B(x + 2) = \dfrac{(x + 2)(x - 5)}{x^2 -3x - 10} = 1; \tag 3$
this may be written
$(A + B)x + (2B - 5A) = 1, \tag 4$
whence
$A + B = 0 \Longrightarrow B = -A, \tag 5$
so that
$-7A = 2(-A) - 5A = 2B -5A = 1 \Longrightarrow A = -\dfrac{1}{7}, \; B = \dfrac{1}{7}; \tag 6$
we check:
$\dfrac{A}{x + 2} + \dfrac{B}{x - 5} = -\dfrac{1}{7}\dfrac{1}{x + 2} + \dfrac{1}{7}\dfrac{1}{x - 5} = \dfrac{1}{7} \left ( \dfrac{1}{x - 5} - \dfrac{1}{x + 2} \right )$
$= \dfrac{1}{7}\left ( \dfrac{x + 2}{x^2 -3x - 10} - \dfrac{x - 5}{x^2 -3x - 10} \right ) = \dfrac{1}{7}\dfrac{7}{x^2 -3x - 10} = \dfrac{1}{x^2 -3x - 10}. \tag 7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$ using Basel Problem sum I have this series I need to calculate:
$$\sum_{n=1}^\infty \frac{n+1}{n(n+2)^2}$$
I've been trying to simplify the fraction via partial fraction decomposition, and I know I somehow need to make use of the Basel Problem, meaning that $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ , but I don't know how to adress the movement of the index $n$ by $2$ in my original problem.
Any tips of how to proceed from here?
| Write (for example using Partial-fraction decomposition)
$$\frac{n+1}{n(n+2)^2} = \frac{1}{2(n+2)^2}+\frac{1}{4}\left(\frac{1}{n}-\frac{1}{n+2}\right).$$
The second bracket summed is a telescoping sum, specifically
$$\sum_{n=1}^k\frac{1}{n}-\frac{1}{n+2}=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\dots+\frac{1}{k}-\frac{1}{k+2},$$
which after canceling terms gives
$$\sum_{n=1}^k\frac{1}{n}-\frac{1}{n+2}=1+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2} \to \frac{3}{2}\,\, \text{as }k\to \infty.$$
For the first expression, notice
$$
\sum_{n=1}^\infty\frac{1}{(n+2)^2}=\sum_{n=3}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{n^2}-1-\frac{1}{4}.
$$
Can you put these together and finish?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral roots of cubic equation $x^3-27x+k=0$ The number of integers $k$ for which the equation $x^3-27x+k=0$ has atleast two distinct integer roots is
(A)$1$
(B)$2$
(C)$3 $
(D)$4$
My Attempt: The condition for cubic $x^3+ax+b=0$to have $3$ real roots happens to be $4a^3+27b^2\leq0$. But how to go about finding condition for integer roots.
| If $x=b$ is one of the solutions
$$k=27b-b^3$$
Now if $b$ is a repeated root and $c$ is the third one,
$0=b+b+c\iff c=-2b$
$$\implies x^3-27x+k=(x-b)^2(x+2b)=x^3+x(2b+b^2)-2b^3$$
$\implies b^2+2b=-27\iff b^2+2b+27=0$ which does not have an integer solution.
So, we can not have repeated roots.
The rest two solutions will be available from the quadratic equation $$0=\dfrac{x^3-27x-(b^3-27b)}{x-b}=x^2+bx+b^2-27$$
As $x$ is an integer, the he discriminant must be perfect square i.e.,
$b^2-4(b^2-27)=108-3b^2=3(36-b^2)=D$(say)
$\implies36-b^2\ge0\iff b^2\le36\implies b\le6$
Also $3$ must divide $b$ to keep $D$ perfect square
So, $b\in[0,\pm3,\pm6]$
Clearly, $D$ is perfect square only for $b=\pm6$ .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of D.
The complex numbers $1+i$ and $1+2i$ are both roots of the equation $x^5-6x^4+Ax^3+Bx^2+Cx+D=0$, where $A, B, C, D \in R$ Find the value of D.
My attempt: The given equation will have 5 roots (distinct or undistinct) since it is a polynomial equation of degree 5. The coefficients are all real. Since $1+i$ and $1+2i$ are two roots of the given equation, their conjugates are also the roots of this equation. Therefore $1-i$ and $1-2i$ are also the roots of this equation. Therefore the L.H.S can be factorized as $x^5-6x^4+Ax^3+Bx^2+Cx+D=\{x-(1+i)\}\{x-(1-i)\}\{x-(1+2i)\}\{x-(1-2i)\}Q(x)=(x^4-4x^3+11x^2-14x+10)Q(x)$
,where $Q(x)$ is a polynomial of degree 1 are it gives the unknown root of the given equation. We find $Q(x)$ is $x-2$ by long division. Multiplying the other factor by $(x-2)$ we get the constant term is -20. Now $D$ is equal to the constant term. Therefore $D$ is equal to -20.
Am I correct? Is there any other way to find $D$?
| Hint:
Use Vieta's formula
$6=1+i+1-i+1-2i+1+2i+t$
where $t$ is the fifth root
Can you complete the solution now?
| {
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Determine the real numbers $a$, $b$, $c$ such that $1$, $\frac1{1+\omega}$ and $\frac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$ I am stuck on this question:
Let $1$, $\omega$ and $\omega^*$ be the cube root of unity.
a. Show that $\dfrac1{1+\omega}=-\omega$ and $\dfrac1{1+\omega^*}=-\omega^*$.
b. Determine the real numbers $a$, $b$, $c$ such that $1$, $\dfrac1{1+\omega}$ and $\dfrac1{1+\omega^*}$ are zeroes of the polynomial $p(z)=z^3+az^2+bz+c$.
c. Hence, find $p(\omega)$ and $p(\omega^*)$.
So I was able to do part a by finding the roots in Cartesian forms, but I am not sure how to approach part b.
| We use the familiar fact that
$$
(x - \alpha) (x - \beta) = x^{2} - (\alpha + \beta) x + \alpha \beta.
$$
For part a, note that $1, \omega, \omega^{*}$ are the three distinct roots of
$$
x^{3} - 1 = (x -1) (x^{2} + x + 1).
$$
Therefore $\omega, \omega^{*}$ are the roots of $x^{2} + x + 1$. Hence their product equals the constant coefficient $1$, so that
$$
\omega^{*} = \omega^{-1},
$$
and their sum equals $-1$, that is, the negative of the coefficient of $x$, so that
$$
\omega + \omega^{*} = -1,
$$
and your formulas follow.
For b, you may note that because of the above formulas,
$$
\frac{1}{1 + \omega} = - \omega,
\quad\text{and}\quad
\frac{1}{1 + \omega^{*}} = - \omega^{*}
$$
are the roots of $x^{2} - x + 1$, and so
$$
1, - \omega, - \omega^{*}
$$
will be the roots of
$$
(x - 1) (x^{2} - x + 1)
=
x^{3} - 2 x^{2} + 2 x - 1.
$$
| {
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"url": "https://math.stackexchange.com/questions/3175872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Numbers of roots in Quadratic Equations If $a,b,c,d$ are real numbers, then show that the equation $$(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$$ has at least two real roots.
| Adding the discriminants of the factors we get $(a^2+12b)+(c^2-4b)+(d^2-8b)=a^2+c^2+d^2.$
If the sum is positive then at least one of discriminants is positive and so at least one of three factors has two distinct roots.
Let $a^2+c^2+d^2=0$ then the equation is $(x^2-3b)(x^2+b)(x^2+2b)=0$ and evidantly has 2 or 4 distinct roots for $b\ne 0$
Thus the equation has less than 2 distinct real roots only when $a=b=c=d=0$
| {
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Alternate complex binomial series sum
Calculation of $\displaystyle \sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}$ is
My Try: Using $$\int^{1}_{0}x^m(1-x)^ndx = \frac{1}{(m+n+1)}\cdot \frac{1}{\binom{m+n}{n}}$$
So $\displaystyle \int^{1}x^{2n-r}(1-x)^r=\frac{1}{2n}\cdot \frac{1}{\binom{2n}{r}}$
Sum convert into $\displaystyle 2n\sum^{2n-1}_{r=1}(-1)^{r-1}r\int^{1}_{0}x^{2n-r}(1-x)^rdx$
$\displaystyle \Longrightarrow 2n \int^{1}_{0}x^{2n}\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r \cdot \bigg(1-\frac{1}{x}\bigg)^rdx$
Could some help me to solve it , Thanks
| $\text{Lemma: } \sum_{k=0}^{n-1} k x^k=\frac{(n-1)x^n+1}{(x-1)}-\frac{(x^n-1)}{(x-1)^2}
$$\text{Proof: }\sum_{k=0}^{n-1} x^k=\frac{x^n-1}{x-1}\implies\sum_{k=0}^{n-1} k x^{k-1}=\frac{n(x-1)x^{n-1}-(x^n-1)}{(x-1)^2}$
$$\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot \frac{1}{\binom{2n}{r}}=\sum^{2n-1}_{r=1}(-1)^{r-1}\cdot r\cdot(2n+1)\int_0^1t^{2n-r}(1-t)^{r}dt$$$$=-(2n+1)\int_0^1t^{2n}\sum_{r=1}^{2n-1}r\Big(1-\frac{1}{t}\Big)^rdt$$$$=-(2n+1)\int_0^1t^{2n}\bigg(\frac{(2n-1){\Big(1-\frac{1}{t}\Big)}^{2n}+1}{\frac{-1}{t}}-\frac{{\Big(1-\frac{1}{t}\Big)}^{2n}-1}{(\frac{-1}{t})^2}\bigg)dt$$
$$=(2n+1)\int_0^1\bigg({(2n-1)t{(1-t)}^{2n}+t^{2n+1}}+t^2{(1-t)}^{2n}-t^{2n+2}\bigg)dt$$
$$=(2n+1)\bigg(\frac{(2n-1)}{(2n+2)(2n+1)}+\frac{1}{2n+2}+\frac{2}{(2n+1)(2n+2)(2n+3)}-\frac{1}{2n+3}\bigg)=\frac{n}{n+1}$$
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Is $x=0$ the inflection point of $f$?
Is $x=0$ the inflection point of $f= \begin{cases} \sin \frac{1}{x}\cdot e^{\frac{-1}{x^2}}, x \neq 0 \\ 0, x=0 \end{cases}?$
I know that the infection point exist when $f''$ changes the sign in this point.I examine that $f$ has $f', f''$ in $x=0$ but from Mathematica I know that $$f''=\begin{cases} -\frac{8 e^{-\frac{1}{x^2}} \left(\frac{\sin \left(\frac{1}{x}\right)}{x^7}-\frac{5 \cos \left(\frac{1}{x}\right)}{x^6}\right)}{x^3}-\frac{2 \left(\frac{4 e^{-\frac{1}{x^2}}}{x^6}-\frac{6 e^{-\frac{1}{x^2}}}{x^4}\right) \cos \left(\frac{1}{x}\right)}{x^5}-2 e^{-\frac{1}{x^2}} \left(-\frac{10 \sin \left(\frac{1}{x}\right)}{x^8}+\frac{30 \cos \left(\frac{1}{x}\right)}{x^7}+\frac{-\frac{\cos \left(\frac{1}{x}\right)}{x^4}-\frac{2 \sin \left(\frac{1}{x}\right)}{x^3}}{x^5}\right), x \neq 0 \\ 0, x=0 \end{cases}$$ so it is really labor-intensive to calculate $f''$ and then calculate when $f''>/<0$.Do you know any faster way to do this?
| $f$ is of the form
$$\tag1f(x)=\begin{cases}\left(p(\frac1x)\sin\frac1x+q(\frac1x)\cos\frac1x\right)e^{-1/{x^2}},&x\ne0\\0,&x=0 \end{cases} $$
where $p,q$ are polynomials. Show that every function of form $(1)$ is differentiable with derivative also of form $(1)$ (but of course with diffferent polynomials $p,q$).
Also, unless both $p$ and $q$ are the zero polynomial, for every function of form $(1)$, there exist sequences $x_n\to 0^+$ and $y_n\to 0^-$such that $f$ changes sign at the $x_n$ and at the $y_n$. Conclude that such $f$ assumes positive and negative values near $x=0$ in such a way that $f$ neither changes sign at $x=0$ nor has an extremum at $x=0$.
Now with the specific $f$ of your porblem statement, apply this result to $f''$ and conclude that $f$ does not have an inflection point at $x=0$.
| {
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$y_0 \ge 2$, $y_n = y_{n-1}^2 -2$ $\Rightarrow$ $\frac{1}{y_0}+\frac{1}{y_0y_1}+\cdots = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$ $y_0 \ge 2$ and $y_n = y_{n-1}^2 - 2$. Let $S_n = \frac{1}{y_0} + \frac{1}{y_0 y_1}+\cdots + \frac{1}{y_0 y_1 \cdots y_{n}}$. Prove that
$$\lim_{n \rightarrow \infty} S_n = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$$
I don't have any idea to deal with it.
| $\text{let} P_n=y_0 y_1\text{...} y_n$, now I will show that $S_n^2-y_0 S_n+1=\frac{1}{P_n^2}$ by MI.
Base case $n=0$ is obvious, assume $S_k^2-y_0 S_k+1=\frac{1}{P_k^2}$, when $n=k+1$
\begin{align}
S_{k+1}^2-y_0 S_{k+1}+1 &=
\left(S_k+\frac{1}{P_{k+1}}\right)^2-y_0 \left(S_k+\frac{1}{P_{k+1}}\right)+1 \\&=
S_k^2-y_0S_k+1+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}+\frac{1}{P_{k+1}^2} \\ &=
\frac{1}{P_k^2}+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}+\frac{1}{P_{k+1}^2}
\end{align}
Then we need $\frac{1}{P_k^2}+\frac{2 S_k}{P_{k+1}}-\frac{y_0}{P_{k+1}}=0$. Again, I will show it by MI.
Base case $n=0$ is obvious, assume $\frac{1}{P_m^2}+\frac{2 S_m}{P_{m+1}}-\frac{y_0}{P_{m+1}}=0$, when $k=m+1$
\begin{align}
\frac{1}{P_{m+1}^2}+\frac{2 S_{m+1}}{P_{m+2}}-\frac{y_0}{P_{m+2}} &=
\frac{1}{y_{m+2}}\left(\frac{y_{m+2}}{P_{m+1}^2}+\frac{2}{P_{m+1}}\left(S_m+\frac{1}{P_{m+1}}\right)-\frac{y_0}{P_{m+1}}\right) \\&=
\frac{1}{y_{m+2}}\left(\frac{y_{m+2}+2}{P_{m+1}^2}-\frac{1}{P_m^2}\right) \\ &=
\frac{1}{y_{m+2}}\left(\frac{y_{m+1}^2}{P_{m+1}^2}-\frac{1}{P_m^2}\right) \\ &=
0
\end{align}
Here I used assumption in the second line.
If $y_0=2$, $\lim S_n =1$ by geometric sum. Obviously $y_0$ larger give smaller $\lim S_n $, so $S_n$ is bounded by $1$.Also, obviously $S_n$ is increasing. So $\lim S_n$ exist for all $y_0\geq2$. Let $L=\lim S_n$.
Take limit on $S_n^2-y_0 S_n+1=\frac{1}{P_n^2}$, we get $L^2-y_0L+1=0$.
Solving the equation, we have $L=\frac{y_0\pm\sqrt{y_0^2-4}}{2}$
$L=\frac{y_0+\sqrt{y_0^2-4}}{2}$ is impossible for $y_0>2$ as it will larger than 1.
| {
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Integral $\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$ Earlier today I saw this integral around here and gave it a try without success, unfortunately it got taken down so it didn't receive to much attention, but I think it's a nice integral (although it seems quite hard) and finding a closed form it's worth trying.
Evaluate $$I=\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$$
I tried to work with it's sister integral:$$J=\int_0^1 \frac{\ln(1-x)}{1+x^3}dx\Rightarrow I-J=-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^3}dx$$
Via the substitution $\frac{1-x}{1+x}=t$ we get:
$$I-J=-\int_0^1 \frac{(1+t)\ln t}{1+3t^2}dt=-\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n}\ln t dt -\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n+1}\ln t dt $$
$$\int_0^1 x^k dx=\frac{1}{k+1}\overset{\frac{d}{dk}}\Rightarrow \int_0^1 x^k\ln x dx=-\frac{1}{(k+1)^2}$$
$$\Rightarrow I-J=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\sum_{n=0}^\infty \frac{(-3)^n }{(2n+2)^2}=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\frac{\operatorname{Li_2 (-3)}}{12}$$
But the first sum is quite ugly looking, so I don't think it's a great approach to the integral and I'm struggling for $I+J$ too.
I remember that OP tried to do partial fractions such as:
$$I=\frac13 \int_0^1 \frac{\ln(1+x)}{1+x}dx-\frac13 \int_0^1 \frac{(x-2)\ln(1+x)}{x^2-x+1}dx$$
And he applied Feynman's trick for the second integral, yet the computation are unbearable and it didn't even spit out the correct result (hopefully there can be a nice closed form and I would like to see one).
| The 'sister integral' approach works for quite a few other integrals, but I do not know how to proceed in this particular case as well (note, by the way, that you have used the geometric series outside its radius of convergence in your calculations), so here's a sketch of the somewhat laborious brute-force method:
After the first partial fraction decomposition we have $I = \frac{1}{6} \ln^2(2) + \frac{1}{3} K$, where
$$ K = \int \limits_0^1 \frac{(2-x) \ln(1+x)}{1 - x + x^2} \, \mathrm{d} x \, .$$
Now we introduce the sixth root of unity $\alpha \equiv \mathrm{e}^{\mathrm{i} \pi/3} = \frac{1 + \sqrt{3} \mathrm{i}}{2}$. It has the useful properties $\overline{\alpha} = 1- \alpha = - \alpha^2$, $\frac{\alpha}{1+\alpha} = \frac{\mathrm{i} \overline{\alpha}}{\sqrt{3}}$, $\frac{\overline{\alpha}}{1+\alpha} = - \frac{\mathrm{i}}{\sqrt{3}}$ and appears when doing partial fractions once more:
$$ \frac{2 - x}{1 - x + x^2} = \frac{- \alpha}{x - \alpha} + \frac{-\overline{\alpha}}{x - \overline{\alpha}} = 2 \operatorname{Re} \left[\frac{- \alpha}{x - \alpha}\right] \, , \, x \in \mathbb{R} \, .$$
Therefore,
\begin{align}
K &= 2 \operatorname{Re} \left[\alpha \int \limits_0^1 \frac{- \ln(1+x)}{x - \alpha} \, \mathrm{d} x \right] \stackrel{t = x - \alpha}{=} 2 \operatorname{Re} \left[\alpha \int \limits_{-\alpha}^{\overline{\alpha}} \frac{- \ln(1+\alpha) - \ln \left(1 + \frac{t}{1+\alpha}\right) }{t} \, \mathrm{d} t \right] \\
&\hspace{-8pt}\stackrel{s = \frac{-t}{1+\alpha}}{=} 2 \operatorname{Re} \left[\alpha \ln(1+\alpha) \left[\ln(-\alpha) - \ln(\overline{\alpha})\right] + \alpha \int \limits_{\frac{\alpha}{1 + \alpha}}^{-\frac{\overline{\alpha}}{1+\alpha}} \frac{- \ln(1-s)}{s} \, \mathrm{d} s \right] \\
&= \frac{\pi^2}{18} + \frac{\pi \ln(3)}{2 \sqrt{3}} + \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] - \sqrt{3} \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] \, .
\end{align}
The dilogarithm values can now be simplified using the various functional equations. We obtain
\begin{align}
\operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) \right] &= \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{\pi^2}{24} + \frac{1}{8} \ln^2(3) \\
\operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right) \right] &= \frac{5 \pi^2}{72} - \frac{1}{8} \ln^2 (3) \, .
\end{align}
The imaginary parts are a bit harder to compute (related questions are found here and here), but a reasonably nice expression in terms of the trigamma function can be derived:
$$ \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] = - \frac{\pi^2}{18 \sqrt{3}} + \frac{\operatorname{\psi}_1 \left(\frac{1}{3}\right)}{12 \sqrt{3}} \, .$$
Thus we arrive at
$$ K = \frac{1}{4} \ln^2 (3) + \frac{\pi \ln(3)}{2 \sqrt{3}} + \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{12} \operatorname{\psi}_1 \left(\frac{1}{3}\right) $$
and
$$ \boxed{I = \frac{1}{6} \ln^2 (2) + \frac{1}{12} \ln^2 (3) + \frac{\pi \ln(3)}{6 \sqrt{3}} + \frac{1}{6} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{36}\operatorname{\psi}_1 \left(\frac{1}{3}\right)} \, .$$
It is of course up to you whether you consider this a nice result, but I have no idea how to simplify it any further.
| {
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searching for a formula for $T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$ So we have a number $n$, which is a power of two.
And we have the following recursion:
$$ T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$$
I solved some exercises like this, but I have a problem with this one.
I don't see the structure:
for the first recursion I have obviously have
$ T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$
for second recursion :
$ T(n) = 2T(\frac{n}{4}) + T(\frac{n}{8}) + \frac{n}{2} +n$
for third recursion :
$ T(n) = 3T(\frac{n}{8}) + 2T(\frac{n}{16}) + 2\frac{n}{4}+ \frac{n}{2} +n$
for the fourth recursion :
$ T(n) = 5T(\frac{n}{16}) + 3T(\frac{n}{32}) + 3\frac{n}{8}+ 2\frac{n}{4} +\frac{n}{2} +n$
for the fifth recursion :
$ T(n) = 8T(\frac{n}{32}) + 5T(\frac{n}{64}) + 5\frac{n}{16}+ 3\frac{n}{8} +2\frac{n}{4} + \frac{n}{2} +n$
.
.
.
So we see that for $k$ we have :
$T(n) = ?T(\frac{n}{2^k})+ ?T(\frac{n}{2^{k+1}}) + \sum_{i=0}^{k-1} ? \frac{n}{2^i}$
As you can see the prefactors are missing.
| For $n=2^N$ let $F(N)=T(n)$. Then
$$T(n) = T(\frac{n}{2}) + T(\frac{n}{4}) + n$$
implies that
$$F(N) = F(N-1) + F(N-2) + 2^N.$$
which is a linear recurrence with characteristic equation $z^2-z-1=0$, the same of the Fibonacci sequence $1,1,2,3,5,8,13,\dots$. Hence the solution is
$$F(N)=A\varphi^N+B(-\varphi)^{-N}+2^{N+2}$$
where $1<\varphi=\frac{1+\sqrt{5}}{2}<2$, and $A,B$ are constants which depend on the initial terms $F(0)=T(1)$ and $F(1)=T(2)$.
Can you take it from here?
| {
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"url": "https://math.stackexchange.com/questions/3195807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 = 11 - 2*4$
$4 = 37-3*11$
$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$
$1 = 3*37 -10*11$
$11(10) - 37(3) = -1$
$11(90) - 37(27) = -9$
x = 90
That is my answer. But the correct answer should be:
x = 16
| We have $$11x\equiv -9\equiv 28\mod 37$$ so we get
$$x\equiv \frac{28}{11}\equiv \frac{65}{11}\equiv\frac{102}{11}\equiv \frac{139}{11}\equiv \frac{176}{11}\equiv 16\mod 11$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove this combinatorial identity Show that
$${2n \choose n} + 3{2n-1 \choose n} + 3^2{2n-2 \choose n} + \cdots + 3^n{n \choose n} \\ = {2n+1 \choose n+1} + 2{2n+1 \choose n+2} + 2^2{2n+1 \choose n+3} + \cdots + 2^n{2n+1 \choose 2n+1}$$
One way that I did it was to use the idea of generating functions.
For the left hand side expression, I can find 2 functions. Consider;
$$f_1 (x) = \frac{1}{(1-3x)} \\ = 1 + 3^1x + 3^2x^2 + 3^3x^3 + \cdots + 3^nx^n + \cdots \\ f_2(x) = \frac{1}{(1-x)^{n+1}} \\ = {n \choose n} + {n+1 \choose n}x + {n+2 \choose n}x^2 + \cdots + {2n-1 \choose n}x^{n-1} + {2n \choose n}x^n + \cdots + $$
Consider the coefficient of $x^n$ in the expansion of $f_1 (x) . f_2 (x)$. Then the coefficient will be the expression on the left hand side.
Now we further consider 2 functions for the right-hand side expression.
Consider;
$$f_3 (x) = \frac {1}{(1-2x)} \\ = 1 + 2^1x + 2^2x^2 + \cdots + 2^{n-1}x^{n-1} + 2^nx^n + \cdots \\ f_4 (x) = (1+x)^{2n+1} \\= 1 + {2n+1 \choose 1}x + \cdots + {2n+1 \choose n-1}x^{n-1} + {2n+1 \choose n}x^n +\cdots + {2n+1 \choose 0}x^{2n +1} \\ = {2n+1 \choose 2n+1} + {2n+1 \choose 2n}x + {2n+1 \choose 2n-1}x^2 + \cdots + {2n+1 \choose n+2}x^{n-1} + {2n+1 \choose n+1}x^{n} + \\
+ {2n+1 \choose n}x^{n+1} +\cdots + {2n+1 \choose 0}x^{2n +1}$$
Hence the coefficient of $x^n$ is the coefficient of $x^n$ in the expansion of $f_3(x) . f_4(x)$
This is what I managed to do so far. I'm not sure if $f_1(x) .f_2(x) = f_3(x).f_4(x)$. If the two functions are indeed equal, then I can conclude that their coefficient of $x^n$ must be equal, which will immediately answer the question. If they are equal, how do I show that they are?
If the two functions are not equal? How do I proceed to show this question?
Edit: It might not be true that the product of the two functions are equal. I tried substituting $x=0.1, n=1$. Seems like the two values are not equal. How do I proceed with this question?
| Using your functions, consider
$$
3^n f_2(\frac13) = 3^n \frac{1}{(1-\frac13)^{n+1}} = \frac32 (\frac92)^n\\ = {n \choose n}3^n + {n+1 \choose n}3^{n-1} + \cdots + {2n \choose n} + {\color{red}{ {2n +1 \choose n} 3^{-1}+ \cdots}}
$$
and further
$$
2^n f_4 (\frac12) = 2^n (\frac32)^{2n+1} = \frac32 (\frac92)^n \\= {2n+1 \choose 2n+1}2^n + {2n+1 \choose 2n}2^{n-1} + \cdots + {2n+1 \choose n+1} + {\color{red}{ {2n +1 \choose n} 2^{-1}+ \cdots + {2n +1 \choose 0} 2^{-n-1}}}
$$
The two expressions both equal $\frac32 (\frac92)^n$, and the first $n+1$ many terms represent the LHS and RHS of the original question. The terms in red are extra terms: once it is established that these terms also equal, the questions is solved. That is, show that
$$
\sum_{k=1}^{\infty}{2n +k \choose n} 3^{-k} = \sum_{k=1}^{n+1}{2n +1 \choose n+1-k} 2^{-k}
$$
| {
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"source": "stackexchange",
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Number of ways to pick a team of 4 with at least 1 girl and 1 boy from 4 girls and 4 boys. To find the number of ways to pick a team of $4$ with at least $1$ girl and $1$ boy from $4$ girls and $4$ boys, I thought to manually assign the team with $1$ girl and $1$ boy to begin with.
$$G, \quad B, \quad G \textrm{ or } B, \quad G \textrm{ or } B$$
The number of ways to pick $1$ girl and $1$ boy from $4$ girls and $4$ boys is
$${4 \choose 1} \times {4 \choose 1}$$
The number of ways to pick the $2$ remaining team members from $6$ people is
$${6 \choose 2}$$
Using this logic the number of teams with at least $1$ girl and $1$ boy is $4 \times 4 \times 15 = 240$, but the total number of teams with no restrictions is ${8 \choose 4} = 70$, so there is definitely something wrong with my thinking.
How do I arrive to the correct answer with this line of reasoning. I already know you can use the fact that it is $70 - (\textrm{number of teams with all boys or all girls}) = 70 - 2 = 68$, but I would like to understand how to do it the way I thought of.
| The counting
$${4 \choose 1} \cdot {4 \choose 1}\cdot {6 \choose 2}=240$$
is the number of teams with 1 "labelled" girl and 1 "labelled" boy. In order to "unlabel" them split ${6 \choose 2}$ into ${3 \choose 2}{3 \choose 0}+{3 \choose 0}{3 \choose 2}+{3 \choose 1}{3 \choose 1}$ and divide each term by the total numbers of boys and girls:
$$\underbrace{{4 \choose 1}}_{\text{1 lab. girl}} \cdot \underbrace{{4 \choose 1}}_{\text{1 lab. boy}} \cdot\left( \frac{1}{3\cdot 1}\underbrace{{3 \choose 2}{3 \choose 0}}_{\text{2 girls and 0 boys}}+\frac{1}{1\cdot 3}\underbrace{{3 \choose 0}{3 \choose 2}}_{\text{0 girls and 2 boys}}+\frac{1}{2\cdot 2}\underbrace{{3 \choose 1}{3 \choose 1}}_{\text{1 girl and 1 boy}}\right)=68.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $a^3 - b^3 = c! - 18$ does not have a solution
Let $a, b,$ and $c$ be positive integers and $c \gt 6$.
Show that the equation $$a^3 - b^3 = c! - 18$$ does not have a solution for all positive integers $a, b,$ and $c$.
What I have realized so far is that if $a^3 - b^3 = c! - 18$, then it must also be true that $a^3 -b^3 \equiv c! - 18 \pmod n$ for all $n \geq 2$.
That would mean if I can show that there exists an $n$ so that $$a^3 - b^3 \not\equiv c! - 18 \pmod n$$ for all relevant $a, b,$ and $c$ then I have also shown that $$a^3 - b^3 \neq c! - 18$$ for all relevant $a,b,$ and $c$.
The only problem that I have now is that I do not know how to find a suitable $n$
| Consider modulo $3$. When $c$ is sufficiently large, arbitrarily large (in particular, $>2$) powers of $3$ divide $c$. Suppose $c$ is sufficiently large in this sense, and suppose $a,b$ existed so that the equation is satisfied. The RHS is $0$ modulo $3$, so we have $a^3=b^3$ mod $3$, which you can easily check implies $a=b$ mod $3$. If $a=b=0$ mod $3$, then the LHS is divisible by $27$, but the RHS is not (since $3\mid c!/9$ but $3\nmid 2$). Suppose $a=b=1$ mod $3$, then set $a=3n+1$, $b=3m+1$. So $a^3-b^3=27(n^3+n^2-m^3-m^2)+9(n-m)=0$ mod $9$. Again, we have a contradiction with the fact that the RHS is not divisible by $9$. Simlarly, then $a=3n-1$ and $b=3m-1$, we have $a^3-b^3=(3n-1)^3-(3m-1)^3=27(n^3-n^2-m^3+m^2)+9(n-m)=0$ mod $9$, and the conclusion follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$ and $y$, The equations are $x \cos^{3} y+3x \cos y \sin^{2} y =14 $ and $ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $ Consider the system of equations
$$x \cos^{3} y+3x \cos y \sin^{2} y =14 $$
$$ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $$
$1)$ the values of $x$ is /are..
Answer is $ \pm\sqrt 5 $
The number of values of $y$ in $(0,6 \pi)$is
Answer is $6$
$Sin^2 y + 2cos^2 y $ is
Answer is $ \frac95 $
I added both the equation and make whole cube
I subtracted both the equation and make whole cube then I divided these equations to find the value of $ \tan y $ = $ \dfrac{1}{2} $
I divided the first equation with $ cos^2 y $ and plugged the value of $\tan y$ which gives $x= +5\sqrt 5 $. But I also need -$ 5\sqrt 5 $ I have no idea where I go wrong.
| I think your approach is quite good.
\begin{align*}
\frac{x(\sin^3y+3\cos^2y\sin y)}{x(\cos^3y+3\cos y\sin^2y)}&=\frac{13}{14}\\
\frac{\tan^3y+3\tan y}{1+3\tan^2y}&=\frac{13}{14}\\
14\tan^3y-39\tan^2y+42\tan y-13&=0\\
(2\tan y-1)(7\tan^2y-16\tan y+13)&=0
\end{align*}
$\tan y=\frac12$.
$y$ has $6$ values in $(0,6\pi)$.
$\sec^2y=1+(\frac12)^2=\frac{5}{4}$.
$\cos y=\pm\sqrt{\frac45}=\pm\frac{2}{\sqrt5}$.
$\sin y=\tan y\cos y=\pm\frac{1}{\sqrt5}$.
So, we have
\begin{align*}
\pm x\left[\left(\frac{1}{\sqrt5}\right)^3+3\left(\frac{2}{\sqrt5}\right)^2\left(\frac{1}{\sqrt5}\right)\right]&=13\\
x&=\pm5\sqrt5
\end{align*}
$\sin^2y+2\cos^2y=(\frac{1}{\sqrt5})^2+2(\frac{2}{\sqrt5})^2=\frac95$.
| {
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| Let $\tan\beta=t$.
Thus, $t\neq1$, otherwise $\gamma=\frac{\pi}{2},$ which is impossible.
Also, $$0<\beta<\frac{3\pi}{4},$$ which is
$$0<\beta<\frac{\pi}{4}$$ or
$$\frac{\pi}{4}<\beta<\frac{\pi}{2}$$ or $$\frac{\pi}{2}<\beta<\frac{3\pi}{4},$$ which is
$$0<t<1$$ or
$$t>1$$ or
$$t<-1.$$
Consider three cases.
*
*$t>1$.
By AM-GM we obtain:
$$p=\tan\beta\tan\left(\frac{3\pi}{4}-\beta\right)=t\cdot\frac{-1-t}{1-t}=\frac{t^2+t}{t-1}=\frac{t^2+t-2+2}{t-1}=$$
$$=t+2+\frac{2}{t-1}=3+t-1+\frac{2}{t-1}\geq3+2\sqrt{(t-1)\cdot\frac{2}{t-1}}=3+2\sqrt2.$$
The equality occurs for $t-1=\frac{2}{t-1}$ and since $\lim\limits_{t\rightarrow1^+}p=+\infty$ and $p$ is a continuous function on $(1,+\infty)$, we got a range of $p$ in this case: $[3+2\sqrt2,+\infty).$
*$0<t<1$
We obtain:
$$p=-\frac{t(1+t)}{1-t}<0.$$ Also, $$\lim\limits_{t\rightarrow1^-}p=-\infty,$$
$$\lim\limits_{t\rightarrow0^+}p=0$$ and $p$ is a continuous function on $(0,1),$
which says that the range of $p$ in this case it's $(-\infty,0).$
*$t<-1$.
In this case $p<0$, $$\lim_{t\rightarrow-1^-}p=0,$$ $$\lim_{t\rightarrow-\infty}p=-\infty$$ and $p$ is a continuous function on $(-\infty,-1).$
Thus, the range of $p$ in this case it's $(-\infty,0),$ which gives the answer:
$$(-\infty,0)\cup[3+2\sqrt2,+\infty).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $\cos(x+y)+\sin(x-y)=2 \sin(x+\frac{\pi}{4}) \sin(\frac{\pi}{4}-y)$ I spend some time trying to figure out how to prove the following identity:
$$ \cos(x+y)+\sin(x-y)=2 \sin\left(x+\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}-y\right) $$
I tried to use the following identities:
$$ \cos(x+y)=\cos(x) \sin(y) - \sin(x) \sin(y) $$
and
$$ \sin(x-y)=\sin(x) \cos(y) - \cos(x) \sin(y). $$
After that, I wanted to use
$$ \cos(x) = \sin\left(x+ \frac{\pi}{2}\right). $$
Unfortunately I can't reach the correct identity. Is there another way of doing it ? Thank you.
| Note that
$$\sin(x+y)+\sin(x-y)=2\sin x \cos y$$
i.e.
$$\sin(A)+\sin(B)=2\sin \frac{A+B}{2} \cos \frac{A-B}{2}$$
Then you can say
\begin{align}
\cos(x+y)+\sin(x-y)
&= \sin(x+y+\frac{\pi}{2})+\sin(x-y)\\
&= 2 \sin(x+\frac{\pi}{4})\cos(y+\frac{\pi}{4})\\
&= 2 \sin(x+\frac{\pi}{4})\sin(\frac{\pi}{4}-y)
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $2^{2^{2^{\cdot^{\cdot^{2}}}}} \mod 9 = 7$
Prove that $\underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{2016 \mbox{
times}} \mod 9 = 7$
I think that it can be done by induction:
Base:
$2^{2^{2^{2}}} \equiv 2^{16} \equiv 2^8 \cdot 2^8 \equiv 2^4 \cdot 2^4 \cdot2^4 \cdot2^4 \equiv 7^2 \cdot 7^2 \equiv 4 \cdot 4 \equiv 16 \equiv 7$
Assume that it is true for $n$ times.
Let $a_n = \underbrace{2^{2^{2^{\cdot^{\cdot^{2}}}}}}_{n \mbox{
times}}$
$$ a_{n+1} = 2^{a_n} = 2^{9k+7} = 2^7 \cdot 2^{9k} \equiv 128 \cdot 1 \equiv 2 \neq 7 $$
Where did I go wrong?
| Where you fail, is in assuming exponents mod the modulus works. It wouldn't generally. the exponent is mod 6, when doing mod 9. So you get it equivalent to: $$128\cdot 2^{3k}=128\cdot8^k$$ Which then depends on parity of k (because 8's order mod 9 is 2).
| {
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"question_score": "2",
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Need help to understand this solution of $2^mp^2 +1=q^5$ The Question-
Find all triples $(m,p,q)$ where $ m $ is a positive integer and $ p ,
> q $ are primes.
****$2^mp^2 +1=q^5$****
After trying my best to solve this problem, and progressing a little bit, I decided to look up the solution and I am unable to quite understand it.
The solution-
clearly $q,p$ are odd
Let:$A=\sum _{i=0}^4q^i$ ,
$\large 2^mp^2=q^5-1=(q-1)\left ( A \right )$
Let:$d=\gcd(q-1,A)$$\large \Rightarrow d|A=\left (\sum _{i=0}^4(q^i-1)
> \right )+5\Leftrightarrow d|5$
If:$d=5\large \Rightarrow p=5\Rightarrow A=5$ since$ A$ is odd , But
$A>1+1+1+1+1=5$
then $d = 1$ $\large \Rightarrow q-1=2^m,A=p^2$ , Assume that $
> m\geq3$ $\large \Rightarrow q\equiv1[8]\Rightarrow p^2=A\equiv5[8]$
Which is impossible , so $m<3$
If:$m=1 \Rightarrow q=3,p=11$ $m=2$ $\large \Rightarrow q=5\Rightarrow
> p^2=781$
Therefore the unique solution is $(1,11,3)$
I have understood upto the part where $d$ divides 5, hence $d=1$ or $5$ but I don't understand how the solution progresses after that.
Could someone give me an explanation behind this proof, and also the thought process of the solution so that I can solve questions like this on my own
| I'm going to add some words to the solution.
$$2^mp^2=(q-1)A\tag1$$
where $A=q^4+q^3+q^2+q+1$.
If:$d=5\large \Rightarrow p=5\Rightarrow A=5$ since$ A$ is odd , But
$A>1+1+1+1+1=5$
If $d=5$, then the right hand of $(1)$ is divisible by $5$, so it follows that $p=5$. Since $A$ is odd, we see that $2^m$ divides $q-1$. Noting that $5$ divides both $q-1$ and $A$, we see that $A=5$. However, this is impossible since $$A=q^4+q^3+q^2+q+1\gt 1+1+1+1+1=5$$
then $d = 1$ $\large \Rightarrow q-1=2^m,A=p^2$ , Assume that $m\geq3$ $\large \Rightarrow q\equiv1[8]\Rightarrow p^2=A\equiv5[8]$ which is impossible , so $m<3$
If:$m=1 \Rightarrow q=3,p=11$ $m=2$ $\large \Rightarrow q=5\Rightarrow p^2=781$
When $d=1$, since $A$ is odd, we see that $q-1=2^m$ and $A=p^2$.
Assume that $m\ge 3$. Then, $q=2^m+1\equiv 0+1\equiv 1\pmod 8$, so $$p^2=A=q^4+q^3+q^2+q+1\equiv 1^4+1^3+1^2+1+1\equiv 5\pmod 8$$
However, there is no integer $p$ satisfying $p^2\equiv 5\pmod 8$.
So, we get $m\lt 3$.
If $m=1$, then $q=2^1+1=3$ and $2p^2=3^5-1$ implies $p=11$.
If $m=2$, them $q=2^2+1=5$ and $4p^2=5^5-1$ implies $p=\sqrt {781}$ which is not an integer.
| {
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Combinatorics generating function question Find the generating function for the sequence $a_n$ which counts the number of bags containing $n$ pieces of fruit in which there is an odd number of apples, at most $1$ banana, and at least $1$ orange. Find $a_{40}$.
I came up with the equation $(x+x^3+x^5...)(1+x)(x+x^2+x^3+...)$ for the three fruits. Simplifying, it becomes $x^2(\frac{1}{1-x^2})(\frac{1-x^2}{1-x})(\frac{1}{1-x})$ and then to $\frac{x^2}{(1-x)^2}$.
Assuming my formula is correct, do I find $a_{40}$ by plugging $40$ into the $x$'s?
| No, $a_{40}$ is the coefficient of $x^{40}$ of the Taylor expansion of $\frac{x^2}{(1-x)^2}$. You can find this by starting $\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+\dots$ and taking the derivative to get $$\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$$
So,
$$\frac{x^2}{(1-x)^2}=\sum_{n=0}^\infty nx^{n+1}$$
What is the coefficient of $x^{40}$?
| {
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How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises.
Theme in the book is mathematical induction. There were examples, where were shown how with mathematical induction prove equations like $(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$.
Now I'm trying to complete exercise where I have to find sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$.
I tried to do it with mathematical induction. Like this:
n=1: $\frac{1}{1+x^2}$
n=2: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}=\frac{1+1+x^2}{(1+x^2)^2}$
n=3: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}=\frac{(1+x^2)^2+2+x^2}{(1+x^2)^3}$
...
And so on (I've calculated til n=5). But I don't see any consistent pattern to evaluate sum of progression.
After that I found formulas of geometrical progression:
$q=\frac{b_{n+1}}{b_n}$ and $S_n=\frac{b_1(1-q^n)}{1-q}$, so I've evaluated:
$q=\frac{\frac{1}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{1}{1+x^2}$
and
$S_n=\left(\frac{1}{1+x^2}\left[1-\left(\frac{1}{1+X^2}\right)^n\right]\right):\left(1-\left[\frac{1}{1+x^2}\right]\right)=\left[\frac{1}{1+x^2}-\left(\frac{1}{1+x^2}\right)^{n+1}\right]\frac{1+x^2}{x^2}=\frac{\left[1-\left(\frac{\sqrt[n+1]{1+x^2}}{1+x^2}\right)^{n+1}\right]}{x^2}$
First of all, I'd like to know how to find sum of geometrical progression with mathematical induction.
Secondly, I'd like to know what is wrong with my evaluations.
| You like to prove $$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$ using Mathematical induction.
The expression is true for $n=1$ because it is simply $q=q$
If true for n, you want to show that it is also true for $n+1$
Note that $$ q+q^2+\dots+q^n+q^{n+1} = \frac{q(1-q^n)}{1-q} + q^{n+1} = $$
$$ \frac {q(1-q^n)+(1-q)q^{n+1}}{1-q}= \frac {q(1-q^{n+1})}{1-q} $$
With $q= \frac {1}{1+x^2}$ the expression under question simplifies to $$\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+…+\frac{1}{(1+x^2)^n}=\frac {(1+x^2)^n-1}{x^2(1+x^2)^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find inverse Laplace transform of : $\ln(\frac{s^2+a^2}{s^2+b^2})$ Question :
Find inverse Laplace transform of :
$$\ln \left(\frac{s^2+a^2}{s^2+b^2}\right)$$
My try :
I'm trying use this identity :
$f(t)=-\frac{\mathcal{L}^{-1}(\frac{dF(s)}{ds})}{t}$
Let $F(s)=\ln(\frac{s^2+a^2}{s^2+b^2})$
Then :
$\frac{dF(s)}{ds}=\frac{2s}{s^2+a^2}-\frac{2s}{s^2+b^2}$
So : $-tf(t)=\mathcal{L}^{-1}(\frac{2s}{s^2+a^2}-\frac{2s}{s^2+b^2})$
$=2\cos (at)-2\cos (bt)$
Is my try correct ?
If any one have simple method plz help me
| Using the change of variable $s=\frac{1}{x}$
$$F\left( s \right)=\ln \left( \frac{{{s}^{2}}+{{a}^{2}}}{{{s}^{2}}+{{b}^{2}}} \right)=\ln \left( \frac{1+{{a}^{2}}{{x}^{2}}}{1+{{b}^{2}}{{x}^{2}}} \right)=\ln \left( 1+{{a}^{2}}{{x}^{2}} \right)-\ln \left( 1+{{b}^{2}}{{x}^{2}} \right)$$
and the power series expansion:
$$\ln \left( 1+u \right)=u-\frac{{{u}^{2}}}{2}+\frac{{{u}^{3}}}{3}-\cdots =\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{u}^{n}}}$$
hence
$$F\left( s \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{\left( \frac{a}{s} \right)}^{2n}}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{\left( \frac{b}{s} \right)}^{2n}}}$$
or
$$f\left( t \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{a}^{2n}}}{n}{{\mathcal{L}}^{-1}}{{\left( \frac{1}{s} \right)}^{2n}}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{b}^{2n}}}{n}{{\mathcal{L}}^{-1}}{{\left( \frac{1}{s} \right)}^{2n}}}$$
so
$$f\left( t \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{a}^{2n}}}{n}\frac{{{t}^{2n-1}}}{\left( 2n-1 \right)!}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{b}^{2n}}}{n}\frac{{{t}^{2n-1}}}{\left( 2n-1 \right)!}}$$
at last
$$\begin{align}
& \frac{1}{2}tf\left( t \right)=\sum\nolimits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{\left( at \right)}^{2n}}}{\left( 2n \right)!}-}\sum\nolimits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{\left( bt \right)}^{2n}}}{\left( 2n \right)!}} \\
& \quad \quad \quad =\left( 1-\cos \left( at \right) \right)-\left( 1-\cos \left( bt \right) \right) \\
\end{align}$$
and you get
$$f\left( t \right)=\frac{2\cos \left( bt \right)-2\cos \left( at \right)}{t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Chebyshev polynomial property
I want to prove inequality (5.13) but I have a problem with (5.16). I have: $$ \sin(n\theta) = \sin\theta \cos(n-1)\theta + \sin(n-1)\theta \cos\theta = $$ $$ = \sin\theta \cos(n-1)\theta + \cos\theta [\sin\theta \cos(n-2)\theta + \sin(n-2)\theta \cos\theta] = $$ $$ = \sin\theta \cos(n-1)\theta + \cos\theta \sin\theta \cos(n-2)\theta + \cos^2\theta \sin\theta \cos(n-3)\theta +... = $$ $$ = \sin\theta [\cos(n-1)\theta + \cos\theta \cos(n-2)\theta + \cos^2\theta \cos(n-3)\theta + \cos^3\theta \cos(n-4)\theta+...] = $$ $$ = \sin\theta [T_{n-1} + T_{1}T_{n-2} + T_{1}^2 T_{n-3} + T_{1}^3T_{n-4}+ ... ] $$ So $$ T_n'(x) = [T_{n-1} + T_{1}T_{n-2} + T_{1}^2 T_{n-3} + T_{1}^3T_{n-4}+ ... = \sum \limits _{i=0} ^{n-1} T_1^i T_{n-1-i} $$ I tried to use fact that $$ T_r(x)T_s(x) = 0.5 (T_{r+s}(x)+T_{r-s}(x)),$$ but I can't prove that $ T'_n(x) = \sum \limits _{i=0} ^{n-1} a_{ik}T_i(x),$ because I can't find $a_{ik}.$
| Note that
\begin{align*}
\frac{\sin n\theta}{\sin\theta}
&=\frac{e^{in\theta}-e^{-in\theta}}{e^{i\theta}-e^{-i\theta}}\\
&=\frac{(e^{i\theta}-e^{-i\theta})[e^{i(n-1)\theta}+e^{i(n-3)\theta}+\dots+e^{-i(n-1)\theta}]}{e^{i\theta}-e^{-i\theta}}\\
&=e^{i(n-1)\theta}+e^{i(n-3)\theta}+\dots+e^{-i(n-1)\theta}\\
&=\begin{cases}
2\cos((n-1)\theta)+2\cos((n-3)\theta)+\dots+2\cos(\theta) & n\text{ even}\\
2\cos((n-1)\theta)+2\cos((n-3)\theta)+\dots+1 & n\text{ odd}
\end{cases}
\end{align*}
So
\begin{align*}
T_n'(x)&=
2T_{n-1}(x)+2T_{n-3}(x)+\dots+\begin{cases}2T_1(x) & n\text{ even}\\T_0(x) & n\text{ odd}\end{cases}\\
&=2\sum_{r=0}^{\lfloor (n-1)/2\rfloor}T_{n-1-2r}(x)+\text{constant term correction}
\end{align*}
and similarly:
\begin{align*}
T_n''(x) &=2\sum_{r=0}^{\lfloor (n-1)/2\rfloor}T_{n-1-2r}'(x)\\
&=4\sum_{r=0}^{\lfloor (n-2)/2\rfloor}(r+1)T_{n-2-2r}(x)+\text{constant term correction}\\
T_n^{(3)}(x) & = 8\sum_{r=0}^{\lfloor (n-3)/2\rfloor}\binom{r+2}{2}T_{n-3-2r}(x)+\text{constant term correction}\\
T_n^{(4)}(x) & = 16\sum_{r=0}^{\lfloor (n-4)/2\rfloor}\binom{r+3}{3}T_{n-4-2r}(x)+\text{constant term correction}\\
\end{align*}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \lim_{x\to \frac{1}{{\sqrt 2}^+}} \frac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\frac{1}{\sqrt{2}}}$
$\displaystyle \lim_{x\to {1\over \sqrt{2}^+}} \dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}$
I have tried substituting $x$ for $\sin \theta$, doing the calculations and ended up with -$2√2$. But the solution provided was $2√2$. Then I tried this question again, but this time used $\cos \theta$ instead of $\sin \theta$ and the answer did match. I don't understand why $x$ as $\sin \theta$ doesn't give the correct result. I have checked all my steps but couldn't find any flaw with $\sin \theta$ as substitution. Can anyone tell me whether $\sin \theta $ a wrong substitution for this question or not?
| Let $x=t+\frac1{\sqrt 2}$ to make
$$A=\dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}=\dfrac{\cos ^{-1}(T)}t\qquad \text{with} \qquad T=2 \left(t+\frac{1}{\sqrt{2}}\right) \sqrt{1-\left(t+\frac{1}{\sqrt{2}}\right)^2}$$
Now, using Taylor expansions around $t=0^+$
$$T=1-4 t^2-4 \sqrt{2} t^3+O\left(t^4\right)$$ Now, using the expansion of $\cos ^{-1}(.)$,we then have
$$\cos ^{-1}(T)=2 \sqrt{2} t+2 t^2+\frac{t^3}{3 \sqrt{2}}+O\left(t^4\right)$$
$$A=2 \sqrt{2}+2 t+\frac{t^2}{3 \sqrt{2}}+O\left(t^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
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